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Physics  

 

 

 

 

 

 

 

 

 

 

 

   

 

 

NEET-UG / AIPMT & JEE (Main) 

Printed at: Repro India Ltd. Mumbai

P.O. No. 32715

For all Agricultural, Medical, Pharmacy and Engineering Entrance Examinations held across India.

10063_11010_JUP

Solutions/hints to Topic Test available in downloadable PDF format at

www.targetpublications.org/tp10063

Salient Features

• Exhaustive coverage of MCQs subtopic wise.

• ‘4024’ MCQs including questions from various competitive exams.

• Includes solved MCQs from MHT CET 2016, NEET P-I and P-II 2016,JEE (Main) 2015 & 16, AIPMT 2015 & Re-Test.

• Various competitive exam questions updated till latest year.

• Concise theory for every topic.

• Neat and authentic diagrams.

• Hints provided wherever relevant.

• Topic test at the end of each chapter.

• Important inclusions: Knowledge bank and Googly questions

PREFACE Target’s “NEET Physics Vol-II” is compiled according to the notified syllabus for NEET-UG & JEE (Main), which in turn has been framed after reviewing various state syllabi as well as the ones prepared by CBSE, NCERT and COBSE. The book comprises of a comprehensive coverage of Theoretical Concepts & Multiple Choice Questions. The flow of content & MCQ’s is planned keeping in mind the weightage given to a topic as per the NEET-UG & JEE (Main) exam. MCQ’s in each chapter are a mix of questions based on theory, numerical and graphical. The level of difficulty of these questions is at par with that of various competitive examinations like CBSE, AIIMS, CPMT, JEE, AIEEE, TS EAMCET (Med. and Engg.), BCECE, Assam CEE, AP EAMCET (Med. and Engg.) & the likes. Also to keep students updated, questions from most recent examinations such as AIPMT/NEET, MHT CET, K CET, GUJ CET, WB JEEM, JEE (Main), of years 2015 and 2016 are exclusively covered. Unique points are represented in the form of Notes at the end of theory section, Formulae are

collectively placed after notes for quick revision and Shortcuts are included to save time of students while

dealing with rigorous questions.

An additional feature of Knowledge Bank is introduced to give students glimpse of various interesting concepts related to the subtopic.

Googly Questions are specifically prepared to develop thinking skills required to answer any tricky or higher

order question in students. These will give students an edge required to score in highly competitive exams.

Topic Test has been provided at the end of each chapter to assess the level of preparation of the student on a competitive level.

We are confident that this book will cater to needs of students of all categories and effectively assist them to achieve their goal. We welcome readers’ comments and suggestions which will enable us to refine and enrich this book further.

All the best to all Aspirants! Yours faithfully Authors

No. Topic Name Page No. 1 Electrostatics 1 2 Current Electricity 108 3 Magnetic Effect of Electric Current 181 4 Magnetism 248 5 Electromagnetic Induction and Alternating Current 302 6 Electromagnetic Waves 371 7 Ray Optics 397 8 Wave Optics 476 9 Interference of Light 494

10 Diffraction and Polarisation of Light 521 11 Dual Nature of Matter and Radiation 554 12 Atoms and Nuclei 596 13 Electronic Devices 664 14 Communication Systems 737

 Note: ** marked section is not for JEE (Main)

1

Chapter 01 : Electrostatics

                               

          

Electrostatics: The study of electricity or electric charges at

rest is known as electrostatics. Charge: i. The property of particles like protons and

electrons which produces electrical influence is called as charge.

ii. It is a scalar quantity. iii. Formula: q = It where q is charge, I is current, t is time. iv. Unit: coulomb in SI system and stat

coulomb in CGS system 1 C = 3 109 stat coulomb stat coulomb is also called electrostatic

unit (e.s.u.) of charge. v. Dimensions: [M0 L0 T1 A1] vi. There are two types of charges: positive

and negative. vii. Like charges repel, unlike charges attract. viii. The electric charge is additive in nature

and is invariant. ix. Accelerated charge radiates energy. x. Charge cannot exist without mass. xi. Charge does not experience any force due

to electric field produced by it. xii. A body can be charged by rubbing or

conduction or induction.

Charging by induction i. The redistribution of charges within a

body due to the presence of a nearby charge is called electrostatic induction.

ii. The charged body induces an opposite charge on the neutral body without actually touching it.

For eg: Attraction of paper bits towards a comb is due to the charging of comb by rubbing it against dry hair. When the charged comb is brought near the paper bits, it induces an opposite charge at the end of paper nearer to the comb resulting in attraction.

Quantisation of charge: i. Charge (q) on a body is always equal to

the integral multiple of electronic charge (e).

q = ne where n = 1,2,3,… ii. Each electron bears a charge equal to

–1.6 10–19 C. Conservation of charges: i. For an isolated system, the net charge

always remains constant. ii. Net charge can neither be created nor be

destroyed in any isolated system. iii. Transfer of charges is possible.

Electric charges and their conservation1.1

Electrostatics01 1.1 Electric charges and their conservation 1.2 Coulomb’s law-force between two point

charges 1.3 Superposition principle, forces between

multiple charges 1.4 Continuous distribution of charges 1.5 Electric field, Electric field lines, Electric

field due to a point charge 1.6 Electric dipole and electric field due to a

dipole 1.7 Torque on a dipole in uniform electric field 1.8 Electric Flux 1.9 Gauss’ theorem and its applications 1.10 Electric potential and **potential difference 1.11 Electric potential due to a point charge, a

dipole and a system of charges

1.12 Equipotential surface 1.13 Electric potential energy of a system of two

point charges and of ** electric dipoles in electrostatic field

1.14 Conductors and insulators, Free and bound

charges inside a conductor 1.15 Dielectrics and electric polarization 1.16 Capacitors and Capacitance 1.17 Capacitance of parallel plate capacitor with

and without dielectric medium between the plates.

1.18 Combination of capacitors in series and

parallel 1.19 Energy stored in a capacitor **1.20 Van de Graaff generator

2 

2

Physics Vol‐II (Med. and Engg.)

Coulomb’s law: i. Force of attraction or repulsion between

two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

F 1 22

q q

r F = C 1 2

2

q q

r

where, q1 and q2 are charges separated by distance r.

ii. In air, C = 0

1

4 in SI system and C = 1 in

CGS system.

In any medium, 0

1C

4 k

in SI system

and C = 1

k in CGS system.

iii. The force between two charges in any medium,

F = 0

1

4 k 1 2

2

q q

r iv. In vector notation, force on q2 due to q1 is

given as,

21F

= 1 2123

0 12

1 q qr

4 k r

where, 12r

is the position vector from q1 to q2.

Force on q1 due to q2 is given as,

12F

= 1 2213

0 21

1 q qr

4 k r

where, 21r

is the position vector from q2 to q1.

v. Coulomb’s force between charges is central force and acts along the line joining the charges.

vi. Coulomb’s force between the two charges is independent of presence of other charges in the surrounding.

Dielectric constant:

Dielectric constant is defined as the ratio of permittivity of any medium () to the permittivity of free space (0).

i.e., k = 0

= k0 where, 0 = 8.85 1012 C2/Nm2 and

0

1

4= 9 109 Nm2/C2 or farad metre–1 (Fm–1)

Superposition principle: i. Total force acting on a given charge due

to number of charges is the vector sum of the individual forces acting on that charge due to all the charges.

ii. Consider number of charges q1, q2, q3…. are applying force on a charge q

Net force on q will be

net 1 2 n 1 nF F F .... F F

iii. The magnitude of the resultant of two

electric forces is given by, 2 2

net 1 2 1 2F = F + F + 2FF cosθ

and 2

1 2

F sinθtanα =

F + F cosθ

iv. The force between any two charges is not affected by the presence or absence of other charges.

Forces between multiple charges: i. Principle of superposition is used to

calculate electric force on a charge due to other charges in the vicinity.

ii. For N point charges q1, q2, …. , qN located

at positions 1 2 Nr , r ,.... r

with respect to

origin respectively, total force 1F

experienced by charge q1 due to all other charges is given by,

1 12 13 14 1NF F F F ........ F

iii. Using vector form of Coulomb’s law,

1 N1 2 1 31 1 2 1 33 3 3l N

01 N1 2 1 2

1 r r r r r rF q q q q ........ q q

4r r r r r r

F2

F1

Fnet

Coulomb’s law force between twopoint charges

1.2 Superposition principle, forcesbetween multiple charges

1.3

q

r2 r3

q1

q2q3

qn–1

qn

r1

rn–1

rn

3

Chapter 01 : Electrostatics

Continuous distribution of charges: i. A system of closely spaced electric

charges form a continuous charge distribution.

ii. On macroscopic level, quantisation of charges is ignored. For a charged body with reasonable size, its charge distribution is treated as continuous.

iii. The continuous distribution can be categorized as linear, surface and volume charge distribution.

Linear charge density: i. When charge is distributed along a line,

charge distribution is called linear.

ii. Linear charge density, q

L

where, L is length of rod. iii. Unit: coulomb metre–1 (Cm–1) iv. Dimensions: [M0 L–1 T1 A1] Surface charge density: i. When charge is distributed over a surface,

charge distribution is called surface charge distribution.

ii. Surface charge density,

q

A

where, A is surface area. iii. Unit: coulomb metre–2 (Cm–2) iv. Dimensions: [M0 L–2 T1 A1] Volume charge density: i. When charge is distributed over the

volume of an object, it is called volume charge distribution.

ii. Volume charge density,

q

V

where, V is volume. iii. Unit: coulomb metre–3 (Cm–3) iv. Dimensions: [M0 L–3 T1 A1] Force due to various charge distribution: i. Force on charge q0 due to line charge

distribution,

02

0

q dˆF r

4 r

l

''

where, r is distance between charge element and point under consideration.

r' is unit vector directed from charge element to the point.

ii. Force on charge q0 due to surface charge distribution,

02

0

q dsˆF r

4 r

''

iii. Force on charge q0 due to volume charge distribution is,

02

0

q dvˆF r

4 r

''

Electric field: i. The space around charge in which its

electric force can be experienced is called electric field.

ii. The electric field is a vector quantity introduced as an intermediary between charges.

Charge field charge iii. The electric field is characteristic of

charge or system of charges and independent of the test charge placed at a point.

iv. The electric field is quantified by electric field intensity.

Electric field intensity

E :

i. Electric field intensity at any point in an electric field is defined as the force acting per unit test charge at that point in an electric field.

ii. Electric field intensity is a vector quantity and represents strength of electric field.

iii. Formula: 0

FE

q

where, q0 is test charge

iv. Unit: newton coulomb–1 (NC–1) or volt metre–1 (Vm–1) in SI system. v. Dimensions: [M1 L1 T–3 A1] vi. It is also called as electric field strength or

electric field. vii. In the presence of dielectric, electric field

decreases and becomes 1

k times its value

in free space. Electric field due to a point charge: i. Consider an isolated point charge q at the

origin. Force experienced by test charge q0

at distance r

is,

02

0

1 qqˆF r

4 r

ii. The electric intensity at the point is given

as,

0

FE

q

2

0

1 qˆE r

4 r

Continuous distribution of charges 1.4

Electric field, Electric field lines,Electric field due to a point charge

1.5

4 

4

Physics Vol‐II (Med. and Engg.)

iii. The magnitude of E

follows inverse square law.

iv. The electric field due to point charge is spherically symmetric.

Electric field lines: i. The imaginary path along which a free

positive charge moves when placed in an electric field is called as electric lines of force.

ii. They start from a positive charge and end on a negative charge.

iii. The tangent to the line of force at any point gives the direction of the electric

field intensity E

at that point. iv. Two lines of force do not intersect each

other. v. The lines of force are normal to the

surface of a charged conductor at any point.

vi. Lines of force do not pass through a conductor. Hence the electric field inside a conductor is always zero. Lines of force can pass through an insulator.

vii. Electric lines of force are crowded together where the field is strong and widely separated from each other where the field is weak.

viii. The lines of force are under tension and tend to shrink. This explains why two unlike charges attract each other.

ix. The lines of force exert a lateral pressure on one another. This explains why like charges repel each other.

Electric dipole: i. System of equal and opposite charges

separated by a small fixed distance is called dipole.

ii. Electric dipole moment p = q(2l) where, 2l = dipole length q = magnitude of either of the charge.

iii. In vector form, ˆp q 2 p

l

The direction of p

is from negative

charge to positive charge. iv. Unit: coulomb metre (Cm) or debye in SI unit. stat C cm in CGS unit. v. Dimensions: [M0 L1 T1 A1] vi. It is a vector. vii. The electric field produced by a dipole is

known as dipole field. viii. The dipole field has a cylindrical

symmetry.

Electric field intensity due to an electric dipole at a point on its axial line (End on position):

i. A line passing through the positive and negative charges of the electric dipole is called the axial line of the electric dipole.

ii. Consider an electric dipole consisting of two point charges ‘– q’ and ‘+ q’ separated by a distance 2l as shown in figure.

The medium between the electric dipole

and the observation point has dielectric constant k.

iii. If 1E

is the electric intensity at P due to charge ‘– q’, then

1 20

1 q ˆE ( i)4 k (r )

l

1E

is represented both in magnitude and

direction by PC .

iv. If 2E

is the electric intensity at P due to charge ‘+q’, then

2 20

1 q ˆE i4 k (r )

l

2E

is represented both in magnitude and

direction by PD . v. Resultant intensity, E = E2 – E1

= 0

q

4 k 2 2

1 1i

(r ) (r )

l l

= 2 2 20

2(q 2 )ri

4 k (r )

l

l

2 2 2

0

1 2prE i

4 k (r )

l

vi. If dipole is of very small length, i.e.,

l < < r, 3

0

1 2pE i

4 k r

For vacuum E = 3

0

1 2p

4 r

vii. The direction of electric field intensity due to electric dipole at a point on its axial line is always along the direction of the dipole moment.

E at a point on the axial line

Axial Line Y

XC P DB OA

r

+ q – q 1E

2E

E

ll

Electric dipole and electric field dueto a dipole

1.6

5

Chapter 01 : Electrostatics

Electric field intensity due to an electric dipole at a point on the equatorial line (Broad side – on position):

i. Equatorial line of an electric dipole is a line perpendicular to the axial line and passing through a point mid-way between the charges of the dipole.

ii. Consider an electric dipole consisting of two point charges ‘– q’ and ‘+ q’ separated by distance 2l as shown in figure.

iii. The medium between the electric dipole

and the observation point has dielectric constant k.

iv. If E1 is the electric field intensity at P due to charge ‘– q’, then

E1 = 2 20

1 q

4 k r l

1E

is represented both in magnitude and

direction by PC . v. If E2 is the electric field intensity at P due

to charge ‘+ q’, then

E2 = 2 2

0

1 q

4 k r l

( BP2 = OP2 + OB2)

2E

is represented both in magnitude and

direction by PD .

vi. Resultant intensity E

at P is, E = E1 cos + E2 cos

= 2 2 3/ 2

0

1 q 2

4 k (r )

l

l

2 2cos

r

l

l

= 2 2 3/2

0

1 p

4 k (r ) l

vii. If dipole is of very small length, i.e., l < < r,

E = 3

0

1 p

4 k r

viii. In vector form, 3

0

1 pE

4 k r

Negative sign indicates E

and p

are in opposite direction.

For vacuum E =3

0

1 p

4 r

ix. Using parallelogram law of vectors, E = 2 E1 cos i.e., Eaxial = 2 Eequator Electric field intensity at a general point due

to short electric dipole: i. Consider a short electric dipole with

dipole moment p

placed in vacuum. Let

O be the mid-point of dipole and line OP

makes an angle with p

.

ii. For any general point P, axial line

component of electric field intensity is,

1 30

1 2p cosE

4 r

along PA.

Equatorial line component of electric field

intensity is, 2 30

1 psinE

4 r

along PB.

iii. Resultant electric intensity is,

2

2 2 21 2 3

0

1 pE E E

4 r

(4 cos2 + sin2)

23

0

1 pE 3cos 1

4 r

iv. From the above figure,

tan = 2

1

E

E =

tan

2

= tan–1 1tan

2

v. Special cases: a. Case I: When P lies on the axial

line of the dipole, = 0.

E = 23

0

1 p3cos 1

4 r

E =3

0

1 2p

4 r

p

r

1E

2E

E

ABE2

+ q

P (r, )

O

p sin

p cos

– q

2E

E

at a point on equatorial

D

R P

C

A

B l l

O

+ q – q

E

1E

r2 2r l2 2r l

6 

6

Physics Vol‐II (Med. and Engg.)

Tube of force

tan = tan 0

2

= 0

= 0 b. Case II: When P lies on the

equatorial line of the dipole, = 90.

E = 23

0

1 p3cos 90 1

4 r

E =3

0

1 p

4 r

tan = tan90

2

=

= 90 i. An electric dipole consisting of two charges

+ q and – q separated by distance 2l is held in

a uniform external electric field E

at an angle as shown.

ii. Two equal and opposite forces act on dipole,

| F | | F | qE

Hence net force is zero. iii. Net torque acts on dipole about an axis

passing through the mid point of dipole. = F 2l sin = qE 2l sin = pE sin

iv. In vector form, p E

v. Special cases: a. Case I: When = 0, then = pE sin 0 = 0 The electric dipole is in stable

equilibrium. b. Case II: When = 90, then = pE sin 90

= pE (maximum value) c. Case III: When = 180, then = pE sin 180 = 0 The electric dipole is in an unstable

equilibrium.

vi. Unit: Nm in SI system dyne – cm in CGS system vii. Dimensions: [M1 L2 T–2 A0] Electric Flux: i. The number of electric lines of force

passing through a given area is called as electric flux.

ii. It is a scalar quantity. iii. The electric flux , through a surface

enclosing ‘q’ is, = 0

q

iv. Unit: Nm2/C or V-m in SI system. dyne-cm2/statcoulomb in CGS system. v. Dimensions: [M1 L3 T3 A1] vi. Electric flux is the dot product of electric

intensity vector and surface area vector,

= E.ds

= Eds cos

vii. For a closed body, outward flux is taken to

be positive and inward flux is negative. viii. Electric flux per unit area is called the flux

density. ix. Electric flux is maximum when electric field

is normal to the area ds i.e., d = Eds. x. Electric flux will be minimum when electric

field is parallel to the area i.e., d = zero. Tube of force: i. The number of lines of

force grouped together to form tube like structure is called as tube of force.

ii. A tube of force originating from unit

positive charge is called unit tube of force or Faraday’s tube of force.

iii. Tube of force has the same properties as line of force.

iv. The number of tubes of force originating

from unit positive charge is 1

=

0

1

k.

v. The number of tubes of force originating

from charge q is 0

q

k.

ds

Torque on a dipole in uniform electric field1.7

Electric flux1.8

E

C

+ q

– q

A

F

p

B

– F

Dipole in uniform electric field

2l

7

Chapter 01 : Electrostatics

Tube of induction: i. Tube of force irrespective of permittivity

of medium is called tube of induction. ii. Only one tube of induction originates

from unit positive charge whatever may be the surrounding medium.

Normal Electric Induction (NEI): i. The number of tubes of induction passing

normally through unit area is called as normal electric induction.

ii. Formula: NEI = E = k0E iii. Unit: C/m2 in SI system and stat C/cm2 in CGS system Total Normal Electric Induction (TNEI): i. TNEI is defined as number of tubes of

induction passing normally through a given area.

ii. The number of tubes of induction passing normally through charge ‘q’ is

N = 0

q

k

iii. No. of tubes of induction passing normally through unit area is k0E,

TNEI = k 0 E ds cos

= k 0 E ds

iv. Unit: coulomb in SI system statcoulomb in CGS system

v. Dimension: [M0 L

0 T1 A1] Gauss’ theorem: i. Total flux through a closed surface is

equal to 0

1

times the total charge

enclosed by that surface.

ii. 0

qE ds

iii. Gauss’ theorem holds good for any closed surface irrespective of its shape or size.

iv. The imaginary surface which encloses the charge or charged body is called Gaussian surface.

Application of Gauss’ Theorem: Electric field due to an infinitely long straight

wire: i. Consider an infinitely long thin wire with

uniform linear charge density . Let P1 and P2 be line elements placed at equal distance on either side of origin ‘O’.

ii. The electric field of every point in plane

normal to wire is radial and its magnitude depends only on radial distance r.

iii. Consider a circular closed cylinder of radius r and length l with infinitely long line of charge as its axis as shown in figure.

iv. Contribution of curved surface of cylinder

towards electric flux,

s

E ds

= E (2rl)

where (2rl) is area of curved surface of cylinder.

v. Total electric flux through cylinder, E = E (2rl) Charge enclosed in cylinder q = l.

vi. From Gauss’ theorem, E = 0

q

E (2rl) = 0

l

E = 02 r

vii. The electric intensity is inversely

proportional to the radius r. E 1

r

viii. If > 0, direction of electric field at every point is radially outwards.

If < 0, direction of electric field at every point is radially inwards.

In this case, E

is due to charge on entire wire while charge enclosed by Gaussian surface is only l.

+ + + + + + + + + + + + + + + + + +

E2

EE1

P2P1

r P

O

Gauss’ theorem and its applications 1.9

r ds

+ + + + + + + + +

P

n n n

n n

n n

E

l

8 

8

Physics Vol‐II (Med. and Engg.)

Electric field due to uniformly charged infinite plane sheet:

i. Consider thin infinite charged plane sheet with surface charge density .

ii. Let P be any point at perpendicular distance r from sheet.

iii. Imagine a cylinder of length 2r and cross sectional area ds around P as shown.

iv. Total electric flux over entire surface of

cylinder, E = 2 E ds v. Total charge enclosed by cylinder, q = ds. vi. From Gauss’ theorem,

E = 2 E ds = 0

q

E = 02

vii. E is independent of r

For > 0, E

is uniform, normal and outwards from sheet.

For < 0, E

is uniform, normal and inwards from sheet.

viii. If infinite plane sheet has uniform thickness,

E = E1 + E2 = 02

+ 02

= 0

Electric field due to uniformly charged

spherical shape: i. Consider a thin spherical shell of radius R

with centre O. Let a charge + q be distributed uniformly over the surface of shell.

ii. Consider a point P, such that OP = r, where r is radius of imaginary sphere S with centre O.

iii. For field outside shell (r > R): According to Gauss’ theorem,

0s

qE ds

E · (4r2) = 0

q

E = 2

0

q

4 r

Thus, for a point outside the spherical

shell, field due to uniformly charged shell acts as if entire charge of shell is concentrated at the centre of the shell.

iv. At a point on the surface of the shell (r = R):

E = 2

0

q

4 r

For surface charge density , q = 4R2 ·

E = 2

20

4 R

4 R

= 0

= constant

v. For field inside shell (r < R): As charge inside a spherical shell is zero,

the gaussian surface encloses no charge. The Gauss theorem gives,

E 4r2 = 0

q

= 0

E = 0 Thus, the field due to a uniformly charged

spherical shell is zero at all points inside shell.

vi. Variation of E

vs r

Y

XO r = R

EmaxE

E2

1

r

Distance from centre (r)

r r

+++++

+ + + + +

+ + + + +

+ + + + +

P

n n

nnn

E

Q

E

E1

E2

R

P

S1

ds

+ q O

r

r E

R

E

ds

S2

O

Pn

r

9

Chapter 01 : Electrostatics

Electric potential: i. Electric potential at any point in an

electric field is defined as work done in bringing a unit charge from infinity to that point against the direction of electric field intensity.

ii. It is a scalar quantity. iii. Formula:

V = 0

W

q

V = 0

1

4 k q

r

where ‘V’ is electric potential at a distance ‘r’ from charge ‘q’.

iv. Unit: volt or JC–1 in SI system statvolt in CGS system (Electrostatic unit of electric potential) 1 volt = 1/300 statvolt v. Dimensions: [M1 L2 T3 A1] Potential difference: i. Work done in bringing a unit charge from

one point to another point against the direction of electric intensity is called as potential difference.

ii. If VA and VB are electrostatic potentials at A and B respectively, then potential difference between them is,

ABB A

0

WV V V

q

where, WAB is work done in moving test charge q0 from point A to B.

Relation between electric intensity and

potential difference: i. When the field is not uniform

E = dV

dx

where dV is small change in potential over small distance dx.

ii. When field is uniform E = V

d

where V is potential difference over distance d.

Electrostatic potential due to a point charge:

i. Potential at point A when a charge (q0) is brought from ,

VA = A

0

W

q

Charge +q situated at point O produces electric field.

ii. Due to presence of electric field, force on

test charge is 0q E

. To move charge

against this force without acceleration external force is needed, which equals

0F q E

iii. If q0 is displaced through small distance dl then small work done is given by,

dW = 0q E d

l

= q0Edl = – q0Edr

Negative sign indicates distance r

decreases in the direction of dl iv. Electric field is given by,

E = 2

0

1 q

4 r

020

1 qdW q

4 r

dr = 0

20

qqdr

4 r

v. Work done in moving a charge from to A is,

WA = A

0

0 A

1 qqdW

4 r

Potential at distance rA is,

VA = 0 A

1 q

4 r

vi. Electrostatic potential at any point at distance r from charge + q is,

V = 0

1 q

4 r

vii. For a single charge, 2

1F

r ;

2

1E

r but

1V

r

viii. At r = , V = q

= 0

i.e., electrostatic potential is zero at infinity.

Electric potential due to system of charges:

P

q3

q2

q1

qn

qi

ri

rnr1

r2

r3

A

O A Q Pdl

0F q E

0q E

q0

+ q

r rA

Electric potential and potential difference1.10

Electric potential due to a point charge, a dipole and a system of charges

1.11

10 

10

Physics Vol‐II (Med. and Engg.)

i. Electrostatic potential at point P due to charge q1 at a distance r1 is,

V1 = 0 1

1 q

4 r

ii. Total potential of point P due to the system of charges is given as sum of the potential at that point due to individual charge.

V = V1 + V2 + V3 + ……. + Vn

= 1 2 n

0 1 0 2 0 n

1 q 1 q 1 q...........

4 r 4 r 4 r

= 1 2 n

0 1 2 n

1 q q q...........

4 r r r

V = n

i

i 10 i

1 q

4 r

iii. a. Electrostatic potential due to discrete charge distribution is given by,

V = n

i

i 10i

1 q

4 | r r |

b. For linear charge distribution,

V = 0 L

1 d

4 | r r |

l

'

where, is linear charge density and r is position vector.]

c. For surface charge distribution,

V = 0 S

1 ds

4 | r r |

where is surface charge density d. For volume charge distribution,

V = 0 V

1 dv

4 | r r |

where is volume charge density. Electric potential at any point due to an

electric dipole: i. Two equal and opposite charges + q and

– q separated by distance 2a form an electric dipole.

ii. Potential at point P due to – q

V1 = 0

1 q

4 PA

And potential at point P due to + q,

V2 = 0

1 q

4 PB

iii. Total potential at a point P is given as, V = V1 + V2

= 0

1 q

4 PA

+ 0

1 q

4 PB

=0

q 1 1

4 PB PA

From the figure, PB = r – a cos PA = r + a cos

V = 0

1

4 q

2 2 2

2a cos

r a cos

V = 2 2 2

0

1 p cos

4 (r a cos )

where, p is electric dipole moment = (2a) q iv. Special cases: a. When = 0, cos = 1

Vaxial = 2 2

0

1 p

4 (r a )

For = , cos = – 1

Vaxial = –2 2

0

1 p

4 (r a )

b. For a < < r,

Vaxial = 2

0

p

4 r

c. When = 90, cos 90 = 0 Vequatorial = 0 v. Electric potential varies inversely as the

square of distance from the centre of dipole.

2

1V

r

Equipotential surface: i. Any surface, which has same electrostatic

potential at every point is called an equipotential surface.

ii. All the points are at same potential. So, no work will be done in moving a charge from one point to another point on the surface.

iii. E

is electric field and a test charge q0 is

moved with displacement dx

on surface, then

q0 E

. dx

= 0 or E

. dx

= 0

dx

is along the surface.

Equipotential surface1.12

A B + q– q

N

r

O

P

a a M

11

Chapter 01 : Electrostatics

Properties of equipotential surface: i. No work is done in moving a test charge

over an equipotential surface. As displacement is along the surface, force applied on it must be perpendicular.

ii. The electric field is always at right angle to the equipotential surface.

iii. Region of strong field and region of weak field can be determined with the help of equipotential surface.

From potential gradient,

E = dV

dr

dr = dV

E

For same value of dV, dr 1

E.

Thus, electric field will be stronger where equipotential surfaces are close and electric field will be weaker where equipotential surfaces are far.

iv. Equipotential surfaces gives the direction of the electric field.

v. Equipotential surfaces cannot intersect each other.

Shapes of equipotential surface due to various charge distributions:

Charge distribution

Figure Shape of

Equipotential surface

Point charge

Spherical

Spherical shell of charge (q)

Spherical

Non-conducting sphere of charge (q)

Spherical

Infinite linear charge distribution

Cylindrical

Electric dipole

Plane

Infinite sheet of charge

Plane

Electric potential energy: i. The work done in displacing a positive

charge (q0) from infinity to a distance r from the charge q is called electric potential energy of test charge. It is denoted by U.

ii. Formula:

U = 0

1

4 k0qq

r iii. Unit: joule in SI system erg in CGS system Also expressed in electron volt (eV) 1 eV = 1.6 1019 J iv. Dimensions: [M1L2T2A0] v. It is a scalar quantity. vi. The change in electric potential energy of

the system is equal to the negative of work done by electric forces.

Electric potential energy of a system of two

point charges: i. The electric potential energy of a system

of two point charges is defined as the amount of work done to assemble this system by bringing charges in from an infinite distance.

equipotential surface

++

++

++

+q

spherical shell

Plane equipotential surface

Electric dipole

+ q q

q

Equipotential surface

equipotential surface

+

+++

+ ++

sphere of charge q

+ +q

Equipotential

surface

+

++

+

+ +

+

+

+

q

Sheet

Infinite line of charge q

equipotential surface

Electric potential energy of a system oftwo point charges and of electricdipoles in electrostatic field

1.13

12 

12

Physics Vol‐II (Med. and Engg.)

ii. Let q1 and q2 be two charges brought in at positions P1 and P2 respectively from

infinity such that P1P2 = 12r

as shown. iii. The electric potential at P2 due to q1 is,

V = 1

0 12

1 q

4 r

iv. Work done in carrying charge q2 at P2, W = potential charge

= 1

0 12

q

4 r q2

v. This work is stored in the system of two point charges q1 and q2 in the form of electric potential energy.

U = W = 1 2

0 12

q q

4 r

Electric potential energy of an electric dipole in uniform electric field:

i. The potential energy of a dipole in uniform electric field is defined as the work done in rotating a dipole from a direction perpendicular to the field to the given direction.

ii. U = (W W90) U = pE (1 cos ) pE U = pE cos

U = p.E

iii. For = 180, potential energy is maximum.

For = 90, potential energy is zero. For = 0, potential energy is minimum. iv. The variation of potential energy as a

function of is shown below

Conductors: i. In conductors, electric charges are free to

move throughout the volume. ii. The electric field inside a charged

conductor is zero. iii. The surface of a charged conductor acts as

an equipotential surface. iv. The total charge of a charged conductor

lies on the outer surface of the conductor. v. Electric lines of force are perpendicular to

the surface of a charged conductor at its every point.

vi. If there is a cavity in a conductor or it is hollow, the electric field inside the cavity is zero. This vanishing of electric field inside the cavity of a conductor is known as electrostatic shielding.

vii. The force dF acting on a small element of area dS of the conductor where the charge density is , is given by

dF = 2

02

dS.

viii. Mostly all metals are conductors. Insulators: i. In insulators, charges remain fixed at their

place. Thus insulators do not have free charges.

ii. An insulator may behave in two ways: a. It may not conduct electricity at all

and is called insulator. Mostly non-metals are insulators.

b. It may not conduct electricity through it but on applying electric field, induced charges are produced on its faces. Such an insulator is called dielectric.

Examples: Water, mica. Free and bound charges inside a conductor: i. In conductors, valence shell is filled less

than half. ii. Due to tendency of an atom to have a

filled valence shell, the valence electrons in atom of conductor leave atom and move freely through lattice of conductor.

iii. The positive ions remain fixed in their positions.

iv. The average velocity of free electrons in conductors is zero.

v. On the application of electric field across the conductor, free electrons experience force. They drift against direction of applied field constituting a net flow through the conductor.

+q1 P1 1(r )

12r P2 2(r )

+q2

+q2 (AT INFINITY)

pE Umax

90 180

pE

Conductors and insulators, Free and bound charges inside a conductor

1.14

13

Chapter 01 : Electrostatics

Dielectrics: Dielectrics are (non-conducting) material that

transmit electric effect without conducting. There are two types of dielectrics:

i. Polar dielectrics ii. Non-polar dielectrics

Polar dielectrics: i. It has permanent electric dipole moment

even if the electric field is absent. ii. Net dipole moment of polar dielectric is

zero because polar molecules are randomly oriented in absence of electric field.

iii. In presence of electric field, polar

molecules align themselves in the direction of the field.

Examples: Alcohol, Water, NH3, HCl. Non-polar dielectrics: i. Every molecule has zero dipole moment

in its normal state. ii. When electric field is applied, molecules

become induced electric dipole. Examples: N2, O2, Methane, Benzene. Electric polarisation: i. The process of inducing equal and

opposite charges, in the two faces of dielectric when external electric field is applied is called electric polarisation. It is expressed in Cm2.

ii. Electric field gets modified in presence of

dielectric, E = E Ei Where E is induced field, E is main field. Ei is field due to dielectrics Dielectric constant: i. Dielectric constant of dielectric medium is

the ratio of the strength of the applied electric field to the strength of the reduced value of the electric field on placing dielectric between the plates of the capacitor.

ii.

electric field between

the plates with airEk

electric field betweenE

the plates with medium

iii. k is called as relative permittivity of the material or SIC (specific inductive capacitance) and k > 1.

Capacitor: i. Capacitor is a device which increases

storing capacity of a conductor at a relatively low potential.

OR An arrangement which consists of two

conductors separated by a dielectric medium such that one of the plate is positively charged and other is connected to the ground is called as capacitor.

It is also called as condenser. ii. Principle of a capacitor: The capacity of

a charged conductor increases if another conductor connected to earth is kept near it.

iii. Types of capacitor: a. Parallel plate capacitor:

C = 0k A

d

Cm = k Cair b. Spherical capacitor:

C = 4k01 2

2 1

r r

r r

c. Cylindrical capacitor:

C = 0

2

1

2 k

r2.303log

r

l

+ + + + + + +

+ ++

+++

++

+

+

r2 r1

+++++

+++++

r2r1

l

E

Ei

++++++

–––––

– +– + – +

– + – +

– +

– + – +– +– +

– ++

– + –

– +–

+ +–

++ –

– –

Dielectrics and electric polarisation1.15

Capacitors and Capacitance 1.16

14 

14

Physics Vol‐II (Med. and Engg.)

iv. A capacitor acts as a small reservoir of energy.

v. Applications of capacitors: a. For tuning in radio circuit. b. For smoothing rectified current in

power supplies. c. For eliminating sparkling of points,

when they are open or close in an ignition system of automobile engine.

d. For storage of large amount of charge in nuclear reactors.

Capacitance:

i. It is defined as the amount of charge required to raise the potential through 1 V.

OR The ability of a conductor to store charge

is called as capacitance of conductor. OR

The ratio of charge on a conductor to its corresponding potential is called capacitance of conductor.

ii. It is a scalar quantity.

iii. Formula: C = q

V

iv. Unit: farad in SI system statfarad in CGS system (i.e., electrostatic unit of capacity) 1farad = 9 1011 statfarad 1 F = 106 F 1 nF = 10–9 F 1 pF = 1012 F v. Dimensions: [M1 L2 T4 A2] vi. Capacitance of a conductor is small and

limited. vii. If a conductor is placed near a charged

conductor, the potential of charged conductor decreases and hence it can store more charge i.e., nearness of a conductor increases the capacitance of a charged conductor.

viii. In a capacitor, it is possible to deposit still larger charges on first plate in presence of induced negative charge on the second plate.

Capacitance of parallel plate capacitor

without dielectric medium: i. Two plates of area A, kept parallel to each

other at a distance d apart and connected to battery.

ii. Charge + q will appear on one plate which will induce a charge q on another plate.

iii. Since d is small, electric field between plate is uniform. The medium between plate is air or vacuum.

iv. The flux through PR and QS of gaussian

surface will be zero as electric field and area are perpendicular to each other.

The flux through RS is given by,

= E.ds

= Eds cos EA

From Gauss’ law,

EA = 0

q

or E =

0

q

A

v. Potential difference between the plates is given by,

E = V

d

V = Ed =

0

qd

A

vi. C = q

V=

0

q

(qd / A) = 0A

d

vii. Capacitance of parallel plate capacitor without dielectric is given by,

C = 0A

d

Capacitance of parallel plate capacitor with dielectric slab between the plates:

i. There are two capacitors with plate area A

and separated by a distance d. ii. The dielectric slab of thickness t and

dielectric constant k is inserted between the plate.

Electric field between the plates is given by,

E = 0

q

k A

A

P

SE0

ER

R

Q

+q

S t d

B Q

q

P

+ + + + + + + + + + + + + +

P+ q

Q

hB

Rd

A

q

ds

E E

S Gaussiansurface

+ + + + + + + + + + + + +

Capacitance of parallel plate capacitorwith and without dielectric mediumbetween the plates

1.17

15

Chapter 01 : Electrostatics

iii. Potential difference between the plate is, V = E0 (d t) + Et

= 0

q

A(d t) +

0

q

kAt

where, 00

qE

A

electric field when k = 1

0

qE

kA

electric field in medium with

dielectric constant k.

V = 0

q

At

(d t)k

V = 0

q

A1

d t 1k

iv. C = q

V

C = 0q A

1q d t 1

k

= 0A1

d t 1k

v. When dielectric slab is completely filled i.e., d = t

C =

0A

kd d kd

= 0Ak

d

= kC0

Capacitors in series: i. In the given figure, three capacitors are

connected in series with each other.

ii. Potential across capacitors are V1, V2 and V3.

iii. Total potential across the capacitors is V, V = V1 + V2 + V3

iv. V1 = 1

q

C, V2 =

2

q

C, V3 =

3

q

C,

S

qV=

C

where, CS is equivalent capacitance in

series.

S

q

C =

1

q

C+

2

q

C +

3

q

C

S

1

C=

1

1

C+

2

1

C +

3

1

C Capacitors in parallel: i. In the given figure three capacitors are

connected in parallel. ii. Potential across each capacitor is V. iii. Charges on each plates of capacitors are

q1, q2, q3. iv. Total charge q = q1 + q2 + q3

q1 = 1C

V, q2 = 2C

V, q3 = 3C

V and q = PC

V

where, CP is equivalent capacitor in parallel.

CP = C1 + C2 + C3 i. When battery is connected across the two

plates of the capacitor, work is done by the battery to charge the capacitor.

ii. Charging of capacitor increases the potential across plates of the capacitor.

iii. The work done by the battery in charging a capacitor is stored in the capacitor in the form of electric potential energy.

iv. Electric potential energy is given by,

U = 1

2

2q

C

+q q +q q +q q

V1 V2 V3

A B E FC D

V

()

=

V

CS

K K+ + ()

+ V

K

C1

C2

C3

+q1

+q2

+q3

q1

q2

q3 qq

Combination of capacitors in seriesand parallel

1.18

Energy stored in a capacitor 1.19

Dielectrics in capacitors serve three purposes: i. To keep the conducting plates from

coming in contact, allowing for smallerplate separations. Thus, resulting inhigher capacitance.

ii. To increase the effective capacitance byreducing the electric field strength.Therefore, we can get the same charge ata lower voltage.

iii. To reduce the possibility of shorting outby sparking during operation at highvoltage.

Knowledge Bank

16 

16

Physics Vol‐II (Med. and Engg.)

Where, q is charge on plates of the capacitor

C is capacitance U is electrical potential energy. v. Energy stored can also be given as,

U = 1

2qV =

1

2CV2

vi. When two charged capacitors are connected together, there is a chance of loss of energy, which can be given by equation,

U = U U =

2

1 2 1 2

1 2

C C V V

2 C C

vii. U is positive so there is always loss of energy. This loss is usually in the form of heat.

i. Van de Graaff generator is an electrostatic

generator which can generate very high potential of the order of 5 106 V. It is used to accelerate charged particles to carry out nuclear reactions.

ii. Principle: It is based on two electrostatic phenomena: a. A sharp pointed conductor has large

charge density. Hence the surrounding air becomes conducting and produces discharge called Corona discharge.

b. When a charged conductor is brought in contact with hollow conductor, it transfers the charge to hollow conductor. The transferred charge resides on the outer surface of the hollow conductor.

iii. It consists of large hollow metallic sphere S mounted on two insulating columns M1 and M2.

iv. A long narrow belt made up of insulating material passes over the pulleys P1 and P2.

v. There are two metal combs C1 and C2, which are called as spray comb (C1) and collector comb (C2).

vi. C1 is at positive potential of about 10,000V by using (HT) source, C2 is connected to sphere.

vii. D is an evacuated accelerating tube having an electrode I at its upper end, which is connected to sphere S.

viii. To prevent leakage of charge from spray, generator is enclosed in steel chamber filled with nitrogen or methane at high pressure.

ix. When spray comb is maintained at high positive potential it produces charges in its vicinity. The positive charges get sprayed on belt due to repulsive action of C1, which are then carried towards C2 by moving belt.

x. Collecting comb C2 is positioned near the upper end of the belt such that pointed ends touch the belt and other end is in contact with the inner surface of metallic sphere.

xi. C2 collects the positive charge and transfers to outer surface of metallic sphere.

xii. As belt goes on moving, accumulation of positive charge on sphere keeps on taking place and raises the potential.

xiii. With increase of charge on the sphere, its leakage due to ionisation of surrounding air becomes faster. When the rate of loss of charge due to leakage becomes equal to the rate at which the charge is transferred to the sphere, the potential of sphere becomes maximum.

xiv. The projectiles such as protons, deuterons are introduced in the upper part of the evacuated accelerator tube. They get accelerated along the length of tube in downward directions and come out with high energy and hit the target with large K.E. and bring about nuclear disintegration.

Applications:

i. It is used to produce high potential of the order of few millions of volt.

ii. It is used to accelerate charged particles such as proton, deuteron.

iii. It is used in nuclear physics for studies.

iv. In medicine, such beams are used to treat cancer. Target

M1 M2 C1

C2

P1

P2

H T

ID

S Metallic comb

Metallic sphere

Insulatingbelt

Motor driven pulley

Steel chamber

Van de Graaff generator1.20

17

Chapter 01 : Electrostatics

1. The electric charge exists on material bodies

which occupy space. Hence electric charge is not a point.

2. The quantisation effect of charge can be

observed only at microscopic level. 3. The charge can be created or destroyed in equal

and unlike pairs only. Examples: pair production, pair annihilation. 4. The electric charge always resides on the outer

surface of the charged conductor. 5. Coulomb’s law is valid only for stationary point

charges and for particle separation ranging from 10–15 m and above.

6. We can use superposition principle for

computing (i) net force (ii) net field (iii) net flux (iv) net potential and (v) potential energy at the observation point P due to any configuration of charges.

7. Two charges Q are at distance r from each

other. If third charge q(= – Q/4) is placed at a distance of (r/2) from each charge, then system of charges will be in equilibrium.

8. At the microscopic level, charge distribution is

discontinuous. 9. Conceptually, continuous charge distribution is

analogous to the continuous mass distribution in mechanics.

10. For a surface charge distribution, electric field

is discontinuous across surface. 11. For a volume charge distribution, electric field

is defined at any point in distribution. 12. The electric field, as per theory, extends upto

infinity. 13. A charge does not experience any force due to

its own electric field. 14. If test charge has significant value, it will

change, value of electric field to be measured. 15. For + q, F

and E

are in same direction i.e.,

perpendicular to surface directing outwards.

For – q, F

and E

are in opposite direction and

E

is perpendicular to surface directing inwards. 16. At a neutral point, E

= 0.

17. If electric field intensity is same both in magnitude and direction throughout then electric field is said to be uniform. Uniform electric field is represented by equispaced parallel lines.

18. Total charge of electric dipole is zero but

electric field of electric dipole is not zero. 19. The atom consisting of positive and negative

charges is not a dipole, as centre of positive and negative charges coincide i.e., 2l = 0.

20. The atom placed in electric field becomes dipole

because positive and negative centres are displaced relative to each other.

21. At all points other than those on axial and

equatorial lines, the field will curve towards or away from the charges.

22. When electric dipole is placed in non-uniform

electric field, p

lies along E

. 23. The value of electric flux is independent of the

distribution of charges and the separation between them inside the closed surface.

24. Gaussian surface cannot pass through any

discrete charge. 25. If wire is not infinitely long, the end effects need

to be considered. 26. For a finite sheet, edge effects need to be

considered. 27. Potential due to a point charge q at its own

location is not defined. 28. Value of V is same at all points at equal

distances from a point charge. Hence potential due to a single charge is said to be spherically symmetric.

29. The potential due to a dipole is axially

symmetric about p

. 32. Electric potential energy of a system of N point

charges is,

U = j N k N

j k

j 1 k 1 0 jkk j

q q1 1

2 4 r

33. The maximum electric field that a dielectric medium can withstand, without breaking down of its insulating property is called its dielectric strength.

34. The capacity of conductor depends on its shape

and size, presence of other charged conductor placed nearly and nature of surrounding medium.

Notes

18 

18

Physics Vol‐II (Med. and Engg.)

35. As the potential of the earth is assumed to be zero, capacity of conductor connected to earth (whatever its shape or charge on it) will be infinite.

36. A capacitor allows ac but blocks dc. 37. Even if the current through capacitor is zero, a

finite amount of energy can be stored in a capacitor.

38. If space between the plates is completely filled with a conductor then, k = and hence C =

39. Series combination is useful, when a single capacitor is not able to tolerate a high potential drop. So it is distributed among number of capacitors.

40. Parallel combination is useful when high capacity is required at low potential.

41. Capacitor actually stores electric energy in the form of electric field between the plates. The net charge on the capacitor is zero.

42. The potential energy stored in the capacitor is independent of the manner in which charge configuration of the capacitor is built up.

43. The potential energy of capacitor lies in the dielectric medium between the plates.

44. The potential energy of capacitor is obtained at the cost of chemical energy stored in the battery used for charging the capacitor.

45. Accelerated particles in Van de Graaff generator are called projectile.

46. Van de Graaff generator is dangerous to handle due to high potential difference.

1. q = It

2. F = 0

1

41 2

2

q q

r

3. net n1 2 n 1F F F ....... F F

Resultant of two electric forces,

netF

= 2 21 2 1 2

F F 2F F cos

2

1 2

F sintan

F F cos

4. 0

FE

q

5. Electric field due to a point charge:

E = 2

0

1 q

4 r 6. p = q (2l)

7. Electric field due to a dipole: For endon (axial) position:

i. E =2 2 2

0

1 2pr

4 k (r ) l

ii. For broad side-on (equatorial) position:

E = 2 2 3/2

0

1 p

4 k (r ) l

iii. For any general point:

E = 23

0

1 p3cos 1

4 k r

8. Angle between resultant electric field intensity and electric field intensity at a point due to axial component:

= tan–1 1

tan2

9. = pE sin 10. = E ds cos = E.ds

11. Gauss theorem:

0

qE.ds

12. Electric field due to:

i. infinitely charged wire, E = 02 r

ii. infinitely charged plane sheet, E = 0

iii. Charged spherical shell, a. For a point outside the shell

E = 2

0

q

4 r

b. For a point on the surface of the

shell, E = 0

c. For a point inside the shell E = 0 13. Electric potential: i. When the field is not uniform,

E = dV

dx

ii. When field is uniform,

E = V

d

14. Potential due to a point charge:

V = 0

1 q

4 r

15. Potential due to a system of charges:

V = n

i

i 10 i

q1

4 r 16. Potential due to an electric dipole:

V = 2 2 2

0

1 pcos

4 (r a cos )

Formulae

19

Chapter 01 : Electrostatics

17. Potential energy: i. For a point charge,

U = 0

0

1 qq

4 k r

ii. For system of two charges,

U = 1 2

0 12

1 q q

4 k r

iii. For electric dipole in uniform field.

U = p E

18. C = q

V 19. For a parallel plate capacitor: i. Without dielectric

C = 0A

d

ii. With dielectric

C = 0A1

d t 1k

20. Capacitors in series:

S

1

C=

1

1

C+

2

1

C +

3

1

C

21. Capacitors in parallel: CP = C1 + C2 + C3 22. Energy stored in capacitor:

U = 1

2CV2 =

1

2qV =

1

2

2q

C

23. Energy loss in capacitor:

U U =

2

1 2 1 2

1 2

C C V V

2 C C

1. When two identical conductors having charges

q1 and q2 are kept in contact and separated later

then each has charge of 1 2q q

2

. If charges are

q1 and q2, then, each has charge 1 2q q

2

. 2. Coulomb’s law is in accordance with the

Newton’s third law of motion.

i.e., q q q q1 2 2 1F F

3. It two charges q1 and q2 are separated by distance r, then distance x between charge q, and null point

x = 2

1

r

q1

q

4. The resultant electric field at a point due to various neighbouring charges is

1 2 3E E E E ........

5. Electric field due to continuous distribution of

charge is, E dE

where dE

is electric field due to a point charge. 6. Direction of the torque is given by right hand

screw rule. 7. For a surface under consideration, electric field

E

is due to both inside as well as outside charges. But term (q) represents only the total charge inside closed surface.

8. By introducing dielectric slab between plates of charged capacitor, capacitance increases, potential difference decreases and charge remains unchanged.

9. If n plates are arranged as shown in figure below, then they constitute (n1) capacitors in parallel. The resultant capacitance of this combination is given by,

C = (n 1) 0A

d

10. If n capacitors each of same capacitance C, when

are connected in series then equivalent

capacitance of combination is CS=C

nand when

are connected in parallel then equivalent capacitance is CP = nC.

11. When two capacitors are connected to each other with wire, the charge on them is redistributed. The ultimate potential of both capacitor is called common potential and is given by,

V = 1 1 2 2

1 2

V C V C

C C

12. Energy spent by the battery in charging the

capacitor = qV,

whereas energy stored = 1

2qV.

13. The total energy stored in series or parallel

grouping of capacitors is sum of energies stored in individual capacitors.

Total energy U = U1 + U2 + U3 + ..... 14. Total energy stored per unit volume of the

capacitor,

u = U 1

volume 2 0E

2

15. The energy stored in parallel plate capacitor is

U = 1

2k0E

2Ad

++++

++++

++++

++++

+ + + +

++++

++++

+

Shortcuts

20 

20

Physics Vol‐II (Med. and Engg.)

1. Which one of the following is the unit of

electric charge? (A) coulomb (B) newton (C) volt (D) coulomb/volt 2. The dimensional formula of electric charge is (A) [M0 L0 T1 A1] (B) [M0 L0 T–1 A1] (C) [M0L0T1A–1] (D) [M0L0T–1A–1] 3. The electric charge always resides (A) at the centre of charged conductor. (B) at the interior of charged conductor. (C) on the outer surface of charged

conductor. (D) randomly all over the charged

conductor. 4. Which among the following is a sure test of

electrification? (A) Attraction (B) Induction (C) Repulsion (D) Conduction 5. A body can be negatively charged by

[AIIMS 1998; C PMT 2009] (A) giving excess of electrons to it. (B) removing some electrons from it. (C) giving some protons to it.

(D) removing some neutrons from it. 6. One metallic sphere A is given positive charge

whereas another identical metallic sphere B of exactly same mass as of A is given equal amount of negative charge. Then

[AMU 1995; R PET 2000; C PMT 2000] (A) mass of A and mass of B still remain

equal. (B) mass of A increases. (C) mass of B decreases. (D) mass of B increases. 7. When a glass rod is rubbed with silk, it

[MP PET 2003] (A) gains electrons from silk. (B) gives electrons to silk. (C) gains protons from silk. (D) gives protons to silk. 8. Two identical conductors of copper and

aluminium are placed in an identical electric fields. The magnitude of induced charge in aluminium will be [AIIMS 2008]

(A) zero (B) greater than in copper (C) equal to that in copper (D) less than in copper

9. When a body is connected to the earth, electrons from the earth flow into the body. This means the body is….. [K CET 2004]

(A) uncharged. (B) charged positively. (C) charged negatively. (D) an insulator. 10. Five balls numbered 1 to 5 are suspended

using separate threads. Pairs (1, 2), (2, 4) and (4, 1) show electrostatic attraction while pairs (2, 3), (4, 5) show repulsion. Therefore, ball 1 must be [BCECE 2015]

(A) neutral (B) metallic (C) positively charged (D) negatively charged 11. Transfer of electric charge can take place in

such quantities which are integral multiples of (A) 1 e.s.u. of charge (B) 1 coulomb (C) 1 micro coulomb (D) 1.6 1019 coulomb 12. A glass rod rubbed with silk is used to charge

a gold leaf electroscope and the leaves are observed to diverge. The electroscope thus charged is exposed to X-rays for a short period. Then [AMU 1995]

(A) the divergence of leaves will not be affected.

(B) the leaves will diverge further. (C) the leaves will collapse. (D) the leaves will melt. 13. Which is bigger, one coulomb or charge on an

electron? (A) One coulomb (B) Charge on electron (C) Both are same (D) None of these 14. When 1014 electrons are removed from a

neutral metal sphere, the charge on the sphere becomes [Manipal MEE 1995]

(A) 16 C (B) –16 C (C) 32 C (D) –32 C 15. Number of electrons in one coulomb of charge

will be [MP PMT/PET 1998; Pb. PMT 1999;

AIIMS 1999; R PET 2001] (A) 5.46 1029 (B) 6.25 1018

(C) 1.6 10–19 (D) 9 1011

16. When 1019 electrons are removed from a

neutral metal plate through some process, the electric charge on it is

[K CET 1999, GUJ CET 2015] (A) –1.6 C (B) +1.6 C (C) 10+19 C (D) 1019 C

Multiple Choice Questions

Electric charges and their conservation1.1

21

Chapter 01 : Electrostatics

17. If a charge on the body is 1 nC, then how many electrons are present on the body?

[K CET 2014] (A) 1.6 × 1019

(B) 6.25 × 109

(C) 6.25 × 1027 (D) 6.25 × 1028

18. A copper sphere of mass 2 g contains about

2 1022 atoms. The charge on the nucleus of each atom is 29e. What fraction of the electrons must be removed from the sphere to give it a charge of +2 C? [BCECE 2014]

(A) 1.08 1011

(B) 2.16 1011 (C) 3.24 1011

(D) 4.32 1011 19. In one gram of a solid, there are 5 1021

atoms. If one electron is removed from every one of 0.01 % of atoms of the solid, charge gained by the solid would be

(A) 0.08 C (B) 0.8 C (C) –0.08 C (D) –0.8 C 20. A conductor has 14.4 10–19 coulomb

positive charge. The conductor has (Charge on electron = 1.6 10–19 coulomb) (A) 9 electrons in excess. (B) 27 electrons in short. (C) 27 electrons in excess. (D) 9 electrons in short. 21. Charge on -particle is [MH CET 2000] (A) 4.8 10–19 C (B) 1.6 10–19 C (C) 3.2 10–19 C (D) 6.4 10–19 C 22. The law governing the force between electric

charges is known as [C PMT 1972; MP PMT 2004]

(A) Ampere’s law (B) Ohm’s law (C) Faraday’s law (D) Coulomb’s law 23. For what order of distance is Coulomb’s law

true? (A) for all distance. (B) distance greater than 10–13 cm. (C) distance less than 10–13 cm. (D) distance equal to 10–13 cm. 24. Dimensions of 0 are (A) [M0 L0 T0 A0] (B) [M0 L–3 T3 A3] (C) [M–1 L–3 T3 A1] (D) [M–1 L–3 T4 A2]

25. In Coulomb’s law, the constant of

proportionality C = 0

1

4has units

(A) N (B) Nm2 (C) Nm2/C2 (D) NC2/m2

26. The magnitude of 0

1

4 is

(A) 9 109 Nm2/C2 (B) 9 109 Nm2/C2 (C) 8.85 1012 Nm2/C2 (D) 8.85 1012 Nm2/C2 27. The ratio of the forces between two small

spheres with constant charge (a) in air (b) in a medium of dielectric constant k is

[MNR 1998] (A) 1 : k (B) k : 1 (C) 1 : k2 (D) k2 : 1 28. When the distance between the charged

particles is halved, the force between them becomes [MNR 1986]

(A) one-fourth (B) half (C) double (D) four times 29. There are two charges +1 microcoulomb and

+ 5 microcoulomb. The ratio of the forces acting on them will be [C PMT 1979]

(A) 1 : 5 (B) 1 : 1 (C) 5 : 1 (D) 1 : 25 30. An electron is moving round the nucleus of a

hydrogen atom in a circular orbit of radius r.

The coulomb force F

between the two is

(Where 0

1k =

4πε) [CBSE PMT 2003]

(A) 2

3

eˆk r

r

(B) 2

3

ek r

r

(C) 2

3

ek r

r

(D) 2

2

eˆk r

r 31. Two point charges placed at a certain distance

r in air exert a force F on each other. Then the distance r at which these charges will exert the same force in a medium of dielectric constant k is given by

[EAMCET 1990; MP PMT 2001] (A) r (B) r / k

(C) r / k (D) r k

Coulomb’s law force between twopoint charges

1.2

22 

22

Physics Vol‐II (Med. and Engg.)

32. Two equally charged, identical metal spheres A and B repel each other with a force F. The spheres are kept fixed with a distance r between them. A third identical, but uncharged sphere C is brought in contact with A and then placed at the mid-point of the line joining A and B. The magnitude of the net electric force on C is

[UPSEAT 2004; DCE 2005] (A) F (B) 3F/4 (C) F/2 (D) F/4 33. Two charges of equal magnitudes and at a

distance r exert a force F on each other. If the charges are halved and distance between them is doubled, then the new force acting on each charge is

[DCE 2010] (A) F / 8 (B) F / 4 (C) 4 F (D) F / 16 34. Identify the wrong statement in the following.

Coulomb's law correctly describes the electric force that

[K CET 2005] (A) binds the electrons of an atom to its

nucleus. (B) binds the protons and neutrons in the

nucleus of an atom. (C) binds atoms together to form molecules. (D) binds atoms and molecules together to

form solids. 35. Two spheres carrying charges +6 μC and

+9 μC, separated by a distance d, experience a force of repulsion F. When a charge of −3 μC is given to both the sphere and kept at the same distance as before, the new force of repulsion is [K CET 2015]

(A) F

3 (B) F

(C) F

9 (D) 3F

36. Two spherical conductors B and C having

equal radii and carrying equal charges in them repel each other with a force F when kept apart at some distance. A third spherical conductor having same radius as that of B but uncharged is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is

[AIEEE 2004] (A) F/4 (B) 3F/4 (C) F/8 (D) 3F/8

37. A total charge Q is broken in two parts Q1 and Q2 and they are placed at a distance R from each other. The maximum force of repulsion between them will occur, when

[MP PET 1990]

(A) 2 1

Q QQ = , Q = Q

R R

(B) 2 1

Q 2QQ = , Q = Q

4 3

(C) 2 1

Q 3QQ = , Q =

4 4

(D) 1 2

Q QQ = , Q =

2 2 38. When air is replaced by a dielectric medium of

constant k, the maximum force of attraction between two charges separated by a distance

(A) decreases k times. (B) remains unchanged. (C) increases k times. (D) increases k2 times. 39. A force F acts between sodium and chlorine

ions of salt (sodium chloride) when put 1 cm apart in air. The permittivity of air and dielectric constant of water are 0 and K respectively. When a piece of salt is put in water, electrical force acting between sodium and chlorine ions 1cm apart is

[MP PET 1995]

(A) F

K (B)

0

FK

(C) 0

F

K (D) 0F

K

40. Dielectric constant of pure water is 81. Its

permittivity will be [C PMT 1984] (A) 7.17 10–10 MKS units (B) 8.86 10–12 MKS units (C) 1.02 1013 MKS units (D) 7.52 10–10 MKS units 41. Two charges placed in air repel each other by

a force of 10–4 N. When oil is introduced between the charges, the force becomes 2.5 10–5 N. The dielectric constant of oil is

[MP PET 2003] (A) 2.5 (B) 0.25 (C) 2.0 (D) 4.0 42. Two charges each of 1 coulomb are at a

distance 1 km apart, the force between them is [D PMT 1999; C PMT 2010]

(A) 9 103 newton (B) 9 10–3 newton (C) 1.1 10–4 newton (D) 104 newton

23

Chapter 01 : Electrostatics

43. Two electrons are separated by a distance of 1Å. What is the coulomb force between them?

[MH CET 2002] (A) 2.3 10–8 N (B) 4.6 10–8 N (C) 1.5 10–8 N (D) 2.8 10–8 N 44. Two charges each equal to 2 C are 0.5 m

apart. If both of them exist inside vacuum, then the force between them is

[C PMT 2001] (A) 1.89 N (B) 2.44 N (C) 0.144 N (D) 3.144 N 45. +2 C and +6 C two charges are repelling each

other with a force of 12 N. If each charge is given –2 C of charge, then the value of the force will be

[C PMT 1979; Kerala PMT 2002] (A) 4 N (Attractive) (B) 4 N (Repulsive) (C) 8 N (Repulsive) (D) Zero 46. The charges on two sphere are +7 C and

–5 C respectively. They experience a force F. If each of them is given an additional charge of – 2 C, the new force of attraction will be

[R PET 2010] (A) F (B) F / 2

(C) F / 3 (D) 2F 47. Two point charges +3 C and +8 C repel

each other with a force of 40 N. If a charge of –5 C is added to each of them, then the force between them will become

[SCRA 1998; JIPMER 2000] (A) –10 N (B) +10 N (C) +20 N (D) –20 N 48. The force between two charges 0.06 m apart is

5 N. If each charge is moved towards the other by 0.01m, then the force between them will become [SCRA 1994]

(A) 7.20 N (B) 11.25 N (C) 22.50 N (D) 45.00 N 49. Two small conducting spheres of equal radius

have charges +10 C and –20 C respectively and placed at a distance R from each other experience force F1. If they are brought in contact and separated to the same distance, they experience force F2. The ratio of F1 to F2 is [MP PMT 2001]

(A) 1 : 8 (B) – 8 : 1 (C) 1 : 2 (D) – 2 : 1

50. Two fixed point charges + 4q and + q units are separated by a distance ‘x’. Where should a third point charge q0 be placed for it to be in equilibrium?

(A) Midway between the charges + 4q and + q.

(B) At a distance 2x from the charge + 4q. (C) At a distance 2x/3 from the charge + 4q. (D) At a distance x/3 from the charge + 4q. 51. In figure two positive charges q2 and q3 fixed

along the y-axis, exert a net electric force in the + x-direction on a charge q1 fixed along the x-axis. If a positive charge Q is added at (x , 0), the force on q1.

i. ii.

[NCERT Exemplar]

(A) shall increase along the positive x-axis. (B) shall decrease along the positive x-axis. (C) shall point along the negative x-axis. (D) shall increase but the direction changes

because of the intersection of Q with q2 and q3.

52. Two positive point charges are 3 m apart and their combined charge is 20 C. If the force between them is 0.075 N, then the charges are

(A) 10 C, 10 C (B) 15 C, 5 C (C) 12 C, 8 C (D) 14 C, 6 C 53. Two charges, each equal to q, are kept at

x = – a and x = a on the x-axis. A particle of

mass m and charge q0 = q

2 is placed at the

origin. If charge q0 is given a small displacement (y << a) along the y-axis, the net force acting on the particle is proportional to

[JEE (Main) 2013] (A) y (B) – y

(C) 1

y (D)

1

y

54. A charge q1 exerts some force on a second

charge q2. If third charge q3 is brought near, the force of q1 exerted on q2 [NCERT 1971]

(A) decreases. (B) increases. (C) remains unchanged. (D) increases if q3 is of the same sign as q1

and decreases if q3 is of opposite sign.

Ox

q3

q2

(x, 0)q1 Q

y

q1

y

x

q3

q2

Superposition principle, forcesbetween multiple charges

1.3

24 

24

Physics Vol‐II (Med. and Engg.)

55. Point charges + 4q , – q and + 4q are kept on the x–axis at point z = 0, z = a and z = 2a respectively.

(A) Only – q is in stable equilibrium. (B) None of the charges are in equilibrium. (C) All the charges are in unstable

equilibrium. (D) All the charges are in stable

equilibrium. 56. If charge q is placed at the centre of the line

joining two equal charges Q, the system of these charges will be in equilibrium if q is

(A) – 4Q (B) – Q/4 (C) – Q/2 (D) + Q/2 57. Three charges each equal to 1 C are placed at

the corners of an equilateral triangle. If force between any two charges is F, then the net force on either will be

(A) 3 F (B) 2 F (C) 3 F (D) F/3 58. Which picture below best represents the

direction of the force on the charge at the origin due to the other two changes shown in figure?

(A) (B) (C) (D) 59. Charges of + 10 C and 20 C are placed as

in figure. The force on a 5 C charge is directed to the right everywhere

(A) in region I. (B) in region II. (C) in region III. (D) in region II and III.

60. Three equal charges are placed on the three corners of a square. If the force between q1 and q2 is F12 and that between q1 and q3 is F13,

the ratio of magnitudes 12

13

F

F is [MP PET 1993]

(A) 1/2 (B) 2

(C) 1/ 2 (D) 2 61. Three charges each of magnitude q are placed

at the corners of an equilateral triangle, the electrostatic force on the charge placed at the center is (each side of triangle is L)

[D PMT 2009]

(A) zero (B) 2

20

1 q

4πε L

(C) 2

20

1 3q

4πε L (D)

2

20

1 q

12πε L

62. Four charges are arranged at the corners of a square ABCD, as shown in the adjoining figure. The force on the charge kept at the centre O is [NCERT 1983; BHU 1999]

(A) zero. (B) along the diagonal AC. (C) along the diagonal BD. (D) perpendicular to side AB. 63. Three charges are arranged as shown in the

figure. The magnitude of the force on q2 due to charge q1 is F21. The magnitude of the force on q2 due to charge q3 is F23. The ratio F21/F23 is

(A) 2/3 (B) 4/3 (C) 2 (D) 3/2 64. Charge q1 = + 6.0 nC is on Y-axis at

y = + 3 cm and charge q2 = – 6.0 nC is on Y-axis at y = – 3 cm. Calculate force on a charge q0 = 2 nC placed on X-axis at x = 4 cm.

(A) – 51.8 jN (B) + 51.8 jN

(C) – 5.118 jN (D) + 5.18 jN

Y

X

+2Q

+Q –Q x

x

+10 C 20 C

I II III

O

C+q

+2qB A

+q

D– 2q

1 m

2 m

3 C

2 C

q3

1 C q2 q1

65

Chapter 01 : Electrostatics

532. The difference in the effective capacitance of two equal capacitors when joined in parallel and series is 3 F. The value of each capacitor is

(A) 1 F (B) 2 F (C) 3 F (D) 4 F 533. Four capacitors each of capacitance 4 F are

connected as shown in the figure. The energy stored in the system is

(A) 5 105 J (B) 5 J (C) 8 105 J (D) 2 J 534. Three identical capacitors are combined

differently. For the same voltage to each combination, the one that stores the greatest energy is [MP PMT 1995; BCECE 2014]

(A) two in parallel and the third in series with it.

(B) three in series. (C) three in parallel. (D) two in series and third in parallel with it. 535. Assertion: A parallel plate capacitor is

charged by a battery. The battery is then disconnected. If the distance between the plates is increased, the energy stored in the capacitor will decrease.

Reason: Work has to be done to increase the separation between the plates of a charged capacitor.

(A) Assertion is True, Reason is True; Reason is a correct explanation for Assertion.

(B) Assertion is True, Reason is True; Reason is not a correct explanation for Assertion.

(C) Assertion is True, Reason is False. (D) Assertion is False, Reason is True. 536. A capacitor of capacitance 5 F is connected

as shown in the figure. The internal resistance of the cell is 0.5 . The amount of charge on the capacitor plate is [MP PET 1997]

(A) 0 C (B) 5 C (C) 10 C (D) 25 C 537. Eight drops of mercury of equal radii and

possesssing equal charges combine to form a big drop. The capacitance of bigger drop as compared to capacitance of each individual drop is

[BCECE 2015] (A) 16 times (B) 8 times (C) 2 times (D) 32 times 538. The charge on a capacitor of capacitance

10 F connected as shown in the figure is [AMU 1995]

(A) 20 C (B) 15 C (C) 10 C (D) zero 539. In the circuit here, the steady state voltage

across capacitor C is a fraction of the battery e.m.f. The fraction is decided by

[AMU (Engg.) 2000]

(A) R1 only. (B) R1 and R2 only. (C) R1 and R3 only. (D) R1, R2 and R3. 540. A combination of capacitors is set up as

shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4 F and 9 F capacitors), at a point distant 30 m from it, would equal:

[JEE (Main) 2016] (A) 360 N/C (B) 420 N/C (C) 480 N/C (D) 240 N/C

C C

C

C

10 V

+ –

5 F 2

1 1

2.5 V

10 F 3

2

2 V

R3

C

R2

R1

+ 8 V

4 F

3 F

9 F

2 F

66 

Physics Vol‐II (Med. and Engg.) 

66

541. Which one statement is correct ? A parallel plate air condenser is connected with a battery. Its charge, potential, electric field and energy are Q0, V0, E0 and U0 respectively. In order to fill the complete space between the plates a dielectric slab is inserted, the battery is still connected. Now the corresponding values Q, V, E and U are in relation with the initially stated as

(A) Q > Q0 (B) V > V0 (C) V < V0 (D) U < U0 542. n identical condensers are joined in parallel

and are charged to potential V. Now they are separated and joined in series. Then the total energy and potential difference of the combination will be

[MP PET 1993] (A) energy and potential difference remain

same. (B) energy remains same and potential

difference is nV. (C) energy increases n times and potential

difference is nV. (D) energy increases V times and potential

difference remains same. 543. 100 capacitors each having a capacity of

10 F are connected in parallel and are charged by a potential difference of 100 kV. The energy stored in the capacitors and the cost of charging them, if electrical energy costs 108 paise per kWh, will be

[MP PET 1996; D PMT 2001] (A) 107 joule and 300 paise.

(B) 5 106 joule and 300 paise.

(C) 5 106 joule and 150 paise. (D) 107 joule and 150 paise. 544. Two capacitors C1 = 2 F and C2 = 6 F in

series, are connected in parallel to a third capacitor C3 = 4 F. This arrangement is then connected to a battery of e.m.f. = 2 V, as shown in the figure. How much energy is lost by the battery in charging the capacitors?

[MP PET 2001]

(A) 22 106 J

(B) 11 106 J

(C) 63210 J

3

(D) 61610 J

3

545. A 40 F capacitor in a defibrillator is charged

to 3000 V. The energy stored in the capacitor is sent through the patient during a pulse of duration 2 ms. The power delivered to the patient is

[AIIMS 2004] (A) 45 kW (B) 90 kW (C) 180 kW (D) 360 kW 546. In air, a charged soap bubble radius of ‘r’ is in

equilibrium having outside and inside pressures being equal. The charge on the drop

is (0 = permittivity of free space, T = surface tension of soap solution)

[MH CET 2014]

(A) 4r2 02T

r

(B) 4r2 04T

r

(C) 4r2 06T

r

(D) 4r2 08T

r

547. If n drops, each of capacitance C and charged

to a potential V, coalesce to form a big drop, the ratio of the energy stored in the big drop to that in each small drop will be

[Assam CEE 2015] (A) n : 1 (B) n4/3 : 1 (C) n5/3 : 1 (D) n2 : 1 548. A particle of mass m and charge +q is midway

between two fixed charged particles each having a charge +q, and at a distance 2L apart. The middle charge is displaced slightly along the line joining the fixed charges and released. The time period of oscillation is proportional to [Assam CEE 2015]

(A) L1/2 (B) L (C) L3/2 (D) L2

C1 C2

C3

+ –

2 V

67

Chapter 01 : Electrostatics

Answers to MCQ's 1. (A) 2. (A) 3. (C) 4. (C) 5. (A) 6. (D) 7. (B) 8. (C) 9. (B) 10. (A) 11. (D) 12. (B) 13. (A) 14. (A) 15. (B) 16. (B) 17. (B) 18. (B) 19. (A) 20. (D) 21. (C) 22. (D) 23. (B) 24. (D) 25. (C) 26. (A) 27. (B) 28. (D) 29. (B) 30. (C) 31. (C) 32. (A) 33. (D) 34. (B) 35. (A) 36. (D) 37. (D) 38. (A) 39. (A) 40. (A) 41. (D) 42. (A) 43. (A) 44. (C) 45. (D) 46. (A) 47. (A) 48. (B) 49. (B) 50. (C) 51. (A) 52. (B) 53. (A) 54. (C) 55. (C) 56. (B) 57. (A) 58. (C) 59. (C) 60. (B) 61. (A) 62. (C) 63. (B) 64. (A) 65. (C) 66. (B) 67. (C) 68. (C) 69. (D) 70. (A) 71. (B) 72. (C) 73. (C) 74. (A) 75. (B) 76. (A) 77. (D) 78. (D) 79. (C) 80. (C) 81. (A) 82. (A) 83. (B) 84. (A) 85. (C) 86. (B) 87. (D) 88. (D) 89. (C) 90. (C) 91. (A) 92. (D) 93. (C) 94. (C) 95. (D) 96. (D) 97. (B) 98. (C) 99. (A) 100. (B) 101. (C) 102. (C) 103. (B) 104. (C) 105. (C) 106. (B) 107. (B) 108. (B) 109. (C) 110. (A) 111. (C) 112. (C) 113. (B) 114. (A) 115. (A) 116. (A) 117. (A) 118. (A) 119. (A) 120. (A) 121. (A) 122. (B) 123. (D) 124. (D) 125. (C) 126. (C) 127. (D) 128. (B) 129. (C) 130. (B) 131. (C) 132. (B) 133. (C) 134. (D) 135. (A) 136. (A) 137. (B) 138. (D) 139. (C) 140. (A) 141. (A) 142. (B) 143. (A) 144. (A) 145. (A) 146. (C) 147. (D) 148. (C) 149. (C) 150. (C) 151. (D) 152. (C) 153. (A) 154. (D) 155. (A) 156. (B) 157. (B) 158. (B) 159. (B) 160. (C) 161. (A) 162. (A) 163. (C) 164. (A) 165. (D) 166. (A) 167. (C) 168. (A) 169. (C) 170. (C) 171. (D) 172. (A) 173. (C) 174. (B) 175. (B) 176. (D) 177. (B) 178. (C) 179. (C) 180. (D) 181. (A) 182. (B) 183. (D) 184. (B) 185. (C) 186. (C) 187. (B) 188. (B) 189. (B) 190. (D) 191. (B) 192. (C) 193. (D) 194. (A) 195. (A) 196. (C) 197. (B) 198. (C) 199. (B) 200. (B) 201. (A) 202. (A) 203. (A) 204. (D) 205. (A) 206. (B) 207. (C) 208. (C) 209. (C) 210. (D) 211. (C) 212. (C) 213. (B) 214. (A) 215. (C) 216. (C) 217. (A) 218. (B) 219. (C) 220. (C) 221. (A) 222. (B) 223. (D) 224. (D) 225. (A) 226. (C) 227. (B) 228. (C) 229. (D) 230. (B) 231. (C) 232. (C) 233. (C) 234. (B) 235. (D) 236. (B) 237. (A) 238. (D) 239. (C) 240. (D) 241. (A) 242. (D) 243. (B) 244. (D) 245. (D) 246. (B) 247. (C) 248. (C) 249. (B) 250. (D) 251. (A) 252. (D) 253. (A) 254. (C) 255. (C) 256. (B) 257. (D) 258. (C) 259. (D) 260. (C) 261. (C) 262. (C) 263. (C) 264. (D) 265. (A) 266. (D) 267. (C) 268. (C) 269. (D) 270. (B) 271. (B) 272. (D) 273. (C) 274. (A) 275. (D) 276. (C) 277. (C) 278. (C) 279. (D) 280. (B) 281. (B) 282. (D) 283. (B) 284. (C) 285. (B) 286. (D) 287. (B) 288. (A) 289. (C) 290. (C) 291. (C) 292. (D) 293. (C) 294. (C) 295. (A) 296. (A) 297. (A) 298. (B) 299. (A) 300. (B) 301. (B) 302. (D) 303. (B) 304. (A) 305. (D) 306. (D) 307. (B) 308. (D) 309. (D) 310. (B) 311. (D) 312. (C) 313. (C) 314. (D) 315. (A) 316. (A) 317. (B) 318. (D) 319. (D) 320. (D) 321. (A) 322. (B) 323. (A) 324. (A) 325. (B) 326. (A) 327. (C) 328. (B) 329. (B) 330. (A) 331. (C) 332. (A) 333. (C) 334. (A) 335. (B) 336. (A) 337. (B) 338. (C) 339. (B) 340. (A) 341. (D) 342. (B) 343. (A) 344. (B) 345. (B) 346. (A) 347. (D) 348. (B) 349. (B) 350. (B) 351. (A) 352. (D) 353. (B) 354. (C) 355. (B) 356. (C) 357. (D) 358. (D) 359. (D) 360. (D) 361. (B) 362. (A) 363. (C) 364. (D) 365. (D) 366. (C) 367. (A) 368. (B) 369. (C) 370. (C) 371. (D) 372. (D) 373. (B) 374. (C) 375. (A) 376. (C) 377. (B) 378. (D) 379. (D) 380. (C) 381. (C) 382. (C) 383. (D) 384. (B) 385. (A) 386. (D) 387. (B) 388. (C) 389. (B) 390. (B) 391. (C) 392. (A) 393. (A) 394. (A) 395. (C) 396. (D) 397. (C) 398. (B) 399. (B) 400. (D) 401. (A) 402. (C) 403. (C) 404. (B) 405. (B) 406. (B) 407. (D) 408. (C) 409. (B) 410. (C) 411. (C) 412. (A) 413. (D) 414. (B) 415. (D) 416. (D) 417. (B) 418. (B) 419. (D) 420. (D) 421. (B) 422. (A) 423. (C) 424. (C) 425. (D) 426. (C) 427. (B) 428. (B) 429. (C) 430. (D) 431. (A) 432. (B) 433. (B) 434. (C) 435. (D) 436. (C) 437. (B) 438. (C) 439. (B) 440. (B) 441. (C) 442. (C) 443. (C) 444. (A) 445. (A) 446. (A) 447. (D) 448. (B) 449. (A) 450. (B) 451. (A) 452. (B) 453. (B) 454. (D) 455. (D) 456. (B) 457. (A) 458. (D) 459. (D) 460. (A) 461. (D) 462. (C) 463. (A) 464. (C) 465. (C) 466. (B) 467. (C) 468. (C) 469. (C) 470. (C) 471. (B) 472. (B) 473. (A) 474. (C) 475. (D) 476. (A) 477. (A) 478. (B) 479. (B) 480. (A) 481. (C) 482. (C) 483. (A) 484. (B) 485. (C) 486. (A) 487. (B) 488. (B) 489. (A) 490. (B) 491. (B) 492. (B) 493. (C) 494. (A) 495. (A) 496. (A) 497. (D) 498. (D) 499. (D) 500. (C) 501. (C) 502. (C) 503. (A) 504. (A) 505. (B) 506. (D) 507. (C) 508. (D) 509. (B) 510. (C) 511. (A) 512. (B) 513. (B) 514. (A) 515. (C) 516. (A) 517. (D) 518. (C) 519. (B) 520. (D) 521. (A) 522. (C) 523. (A) 524. (A) 525. (B) 526. (A) 527. (D) 528. (B) 529. (C) 530. (C) 531. (A) 532. (B) 533. (C) 534. (C) 535. (D) 536. (C) 537. (C) 538. (A) 539. (B) 540. (B) 541. (A) 542. (B) 543. (C) 544. (B) 545. (B) 546. (D) 547. (C) 548. (C)

68 

Physics Vol‐II (Med. and Engg.) 

68

2. Charge = current time = AT = [M0L0T1A1] 5. Excess of electron gives the negative charge

on body. 6. Negative charge means excess of electrons

which increases the mass of sphere B. Whereas positive charge on sphere A is given by removal of electrons.

7. On rubbing glass rod with silk, excess

electrons are transferred from glass to silk. So glass rod becomes positive and silk becomes negative.

8. Since both are metals so equal amount of

charge will be induced in them. 9. When a positively charged body connected to

the earth, electrons flows from earth to body and body becomes neutral.

10. As 2 and 3 repel and also 4 and 5 repel; 2, 3, 4

and 5 must be charged. As 2 and 4 attract, they must be oppositely

charged. As 1 attracts both 2 and 4. This is possible

only for a neutral ball. Hence, 1 is neutral. 12. Charge on glass rod is positive, so charge on

gold leaves will also be positive. Due to X-rays, more electrons from leaves will be emitted, so leaves becomes more positive and diverge further.

13. No. of electrons constituting 1 C,

1919

q 1n 0.625 10

e 1.6 10

14. 14 19q = ne = 10 ×1.6×10

q = 1.6 105 C = 16 C

Electrons are removed, so charge will be positive.

15. 1819

q 1n = = = 6.25×10

e 1.6×10

16. By using Q = ne Q = 1019 1.6 1019

Q = +1.6 C

17. n = q

e=

9

19

1 10

1.6 10

= 6.25 109

18. Q = ne

n = Q

e

n = 6

22 19

2 10

2 10 29 1.6 10

n = 2.16 1011

19. n = 0.01

100 5 1021 = 5 1017

q = – ne = – 5 1017 (–1.6 10–19) = 0.08 C. 20. Positive charge shows the deficiency of

electrons.

Number of electrons 19

19

14.4×10= = 9

1.6×10

21. By using, q = ne q = + 2e = + 3.2 10–19 C

24. F = 0

1

4πε 1 2

2

q q

r

0 = 1

4πF 1 2

2

q q

r

= 1 1 -2

1

[M LT ]

2 2

2

[A T ]

[L ]

= [M–1 L–3 T4 A2]

25. Constant of proportionality C = 0

1

4πε

Unit of 0 = C2 / Nm2

Unit of 0

1

4=

2 2

1

C / Nm =

2

2

Nm

C

{ 4 has no dimensions}

27. 1 2 1 2a b2 2

0 0

q q q qF = ,F =

4πε r k4πε r Fa : Fb = k : 1

28. 2

1F

r ; so when r is halved the force becomes

four times. 29. The same force will act on both bodies

although their directions will be different.

1 2| F | | F |

30. 2 2

2 3

e eˆF = k r = k. r

r r

r

r =r

31. F = F

1 2 1 22 2

0 0

q q q q=

4πε r 4πε r k'

r

rk

Hints to MCQ's

1 2 3 4 5

= repulsion

= attraction

69

Chapter 01 : Electrostatics

32. Initially,

2

2

qF = C

r

where C is constant of proportionality Finally, When uncharged sphere C is brought in

contact with charged sphere A, the sphere C

acquires a charge Q

2due to conduction.

Force on C due to A, 2 2

A 2 2

C(Q / 2) CQF = =

(r / 2) r

Force on C due to B, 2

B 2 2

CQ(Q / 2) 2CQF = =

(r / 2) r

Net force on C, 2

net B A 2

CQF = F F = = F

r

33. 2

2

qF = C

r. If q is halved, r is doubled then

2

2

2 2

qC

Cq2F

4 4 r2r

=

2

2

1 Cq F

16 r 16

34. Nuclear force binds the protons and neutrons

in the nucleus of an atom. 35. F Q1Q2 ( r is same in both cases)

F

F = 1 2

1 2

Q Q

Q Q =

6 9

(6 3) (9 3)

= 3

F = F

3

36. Initially 2

2

QF = C

r (figure A). Finally when a

third spherical conductor comes in contact alternately with B and C then removed, so charges on B and C are Q/2 and 3Q/4 respectively (figure B)

Now force 2

Q 3Q32 4

F = C = Fr 8

37. Q1 + Q2 = Q ….(i) and 1 22

Q QF = C

r ….(ii)

From (i) and (ii), 1 12

CQ (Q Q )F =

r

For F to be maximum 1

dF= 0

dQ

0 = 21 12

1

C dQQ Q

r dQ

0 = Q 2Q1 Q = 2Q1

Q1 = Q

2

Q2 = Q

2

38. 1 22

0

1 q q FF = =

4πε k r k

If F is the force in air, then F is less than F since k > 1.

39. F = 1 22

0

q q1

4 K r

F 1

K

For air K = 1 Let F be electrical force when placed in water

F 1

F K

F

FK

40. = k0 = 81 8.854 10–12 = 7.17 10–10 MKS units

41. 4

a5

m

F 10k = k = = 4

F 2.5×10

42. 2

9 2 32 2

Cq 1F = = 9×10 ×1 × = 9×10 N

r (1000)

43. 2

92

qF = 9×10 ×

r

19 2

9 810 2

(1.6×10 )= 9×10 × = 2.3×10 N

(10 )

44. By using 2

92

qF = 9×10 .

r

F = 6 2

92

(2×10 )9×10 .

(0.5)

= 0.144 N

45. Resultant charges after adding the –2 C be

(–2 + 2) = 0 and (–2 + 6) = + 4 C

F 1 22 2

Cq q 0× 4= = C× = 0

r r

46. 6 6

20

1 (+7×10 )( 5×10 )F =

4πε r

12

20

1 35×10= N

4πε r

A B r

Q Q

A C r/2

Q/2 Q

B r/2

FB FA Q/2

Q

B C r

Q

(A)

Q/2

B Cr

3Q/4

(B)

70 

Physics Vol‐II (Med. and Engg.) 

70

F = 6 6

20

1 (+5×10 )( 7×10 )

4πε r

12

20

1 35×10= N

4πε r

F = F 47. In second case, charges will be –2 C and +3 C

Since F q1q2 i.e., 1 2

1 2

F q q=

F q q

40 3×8

= = 4F 2×3

'

F = –10 N

48. 2

1F

r

2

1 2

2 1

F r=

F r

2

2

5 0.04=

F 0.06

2F =11.25N 49. F q1q2 When both spheres are brought in contact with

each other, each sphere acquires a charge, 10 20

2

= 5 c

1 1 2

2 1 2

F q q 10× 20 8= = =

F q q 5× 5 1

50.

F = 0 02 2

4q×q q×q=

a (x a)

4(x a)2 = a2 2(x – a) = a 2x – 2a = a

3a = 2x a = 2x

3 51. As, net force on charge q1, due to positive

charges q2 and q3 is along + x-axis. Thus, it is an attractive force.

Hence, q1 must be a negative charge.

Now, if positive charge Q is added at (x, 0), it

exerts an attractive force (fqQ) on q1, along + x-axis.

Thus, the force on q1 shall increase along the positive x-axis.

52. F = 0

1

4 1 2

2

q q

r

0.075 = 9

1 22

9 10 (q q )

(3)

q1 q2 = 9

0.075 9

9 10

= 0.075 10–9 = 75 10–12 ….(i) Given, q1 + q2 = 20 10–6 q1 = 20 10–6 – q2 Putting in (i), (20 10–6 – q2) q2 = 75 10–12 20 10–6 q2 – q2

2 = 75 10–12 q2

2 – 20 10–6 q2 + 75 10–12 = 0 q2

2 –1510–6 q2 – 510–6 q2 + 75 10–12 = 0 q2 (q2 – 1510–6)–510–6 (q2 – 1510–6) = 0 q2 = 15 10–6 or q2 = 5 10–6 Two charges are 15 C, 5 C. 53. Fnet = 2 F cos

= 2

2 2

2 kq

2 a y

cos

= 2

2 2

kq

a y

2 2

y

a y

For y << a

2

3

kq yF

a

F y

a

(qo)

x

(xa) +q+4q

xq1

+q2

+y

+x

y

+q3

fnet

f = fnet + fqQ x

+q3

+q2

+Q(x, 0)-q1

+y

–y

+x

y

q

F

qO

Fnet F

q/2

a a

71

Chapter 01 : Electrostatics

54. The force will still remain 1 22

0

q q

4 r according

to the superposition principle. 56. For equilibrium, net force on Q = 0

2 2

CQQ CqQ0

(2x) x

q = – Q/4 57. Each charge experiences two forces inclined at

60 . Therefore,

R = 2 2 oF + F + 2FFcos60 = 23F = 3 F 59. F q1q2 FL (5) (+10) = 50 FR (5) (20) = +100 As FR > FL In I Left or Right

depends on distance. In II Always Left In III Always Right 60.

2

12 20

1 qF =

4πε a and

2

13 20

1 qF =

4πε (a 2)

12

13

F= 2

F

61. In the following figure since A CB| F | | F | | F |

and they are equally inclined with each other, so their resultant will be zero.

62. We put a unit positive charge at O. Resultant

force due to the charge placed at A and C is zero and resultant force due to B and D is towards D along the diagonal BD. For a unit negative charge, resultant force is towards B along BD.

63. F 1 22

q q

r

F21

1× 2

1 and F23

2

3 2

(2)

=

3

2

21

23

F

F =

2

(3 / 2)=

4

3

64. Here, q = 6.0 nC = 6.0 10–9 C 2 a = 6 cm = 6 10–2 m r = 4 cm (on equatorial line) = 4 10–2 m q0 = 2 nC = 2 10–9 C, F = F1 cos + F2 cos

= 2 02

0

1 qq

4 r cos

= 2 9 109 9 9

2 2

6 10 2 10 3

(5 10 ) 5

= 5.184 10–5 N

F

= ˆ51.8j N 65. Net force on B, 2 2

net A CF F F

A 2

15 12F 20 dyne

3

,

C 2

12 20F 15 dyne

4

2 2 2 2net A CF F F (20) (15) 25 dyne

66. FA = force on C due to charge placed at A

6 6

92 2

10 2 109 10 1.8N

(10 10 )

FB = force on C due to charge placed at B

6 6

92

10 2 109 10 1.8N

(0.1)

Net force on C,

2 2 onet A B A BF (F ) (F ) 2F F cos120 1.8N

q2

q3

q1

2 a

a

a

CF

q A

BC q q AF

BFQ

3

0

5

5

4

q1

q2

F2

q0 F1

F1 cos + F2 cos3

+1C – 1C

+2C

10 cm

FB

B A

C

FA

120o

A

FC

FA

3 cm

4 cm

+15 esu

–20 esu+12 esuB C

2 2net A CF F F

101

Chapter 01 : Electrostatics

503. T cos = mg ….(i)

T sin = 0

q

2

….(ii)

Dividing equation (ii) by equation (i),

tan = 0

q

2 mg

504. Taking into account the horizontal component of electric field at point B,

dEx = dE cos

dEx = 2 2 2 20

( d )

4 (r ) r

x x

x x

2 2From figure, cos

r

x

x

dEx = 2 2 3/2

0

d

4 (r )

x x

x

Ex = 2 2 3/20 0

ddE

4 (r )

x

x x

x

Ex = 2 2 3/2

0 0

12 (r ) d

4 2

x x x

Let r2 + x2 = t differentiating w.r.t. t, 2x dx = dt

2 2 3/2

0

(r ) 2 d

x x x = 3/2

2r

t dt

On integrating we get, 2/r

Ex = 04

1

2

2

r

Ex = 04 r

Similarly, EY = Y0

dE dEsin4 r

Electric field at point B, E = 2 2x yE E =

0

2

4 r

tan = y

x

E1

E

= tan1(1) = 45

E

makes an angle of 45 with AB.

505. E = 0

R

kr

= 2 6

0

2 10 5 10

0.4 1

= 8

10

10 10

4 10 1

= 9 109 10–6

= 9 103 N/C 506. Electric field due to charge Q at r = a is,

Ea = 2

kQ

a ….(i)

Consider a shell of thickness dr in the region a r b.

charge on shell, dq = Area = 4 r2d r A

r

total charge in the region a r b is,

q = b

a

dq = 4A

b

a

r dr = 4A b2

a

r

2

q = 2A [b2 a2] Electric field at r = b is,

Eb = 2 2

2

k 2 A b a Q

b

….(ii)

For electric field to be constant in the region a r b we must have, Ea = Eb

From equations (i) and (ii),

2 2

2 2

2 A b a QkQk

a b

2

2 22

Q bQ 2 A b a

a

2 2

2 22

Qb Qa2 A b a

a

2 2

2 22

Q b a2 A b a

a

A = 2

Q

2 a

T cos

T sin

mg

A

T qE =

0

q

2

A

O

dx x

A

2 2r x r

dEx

dE dEy

B

Qa

r dr

102 

Physics Vol‐II (Med. and Engg.) 

102

507. When the particle lies in the plane of the ring where no net force acts on it, then it must lie at the centre of the ring.

Suppose, the particle is displaced slightly from mean position by x (say).

Electric field due to the ring at the new position of the particle is given by

E = 2 2 3/ 2

0

(2 a)

4 (a )

x

x

For x < < a, E = 2

02 a

x ….(i)

(directed away from centre of ring) Now, force acting towards the centre of the

ring is given by F = – qE

F = – 20

q2 a

x ….[using (i)]

ma = – 20

q2 a

x (F = ma)

a = – 20

q

2 ma

x ….(ii)

Also, equation for SHM is given by a = – 2 x ….(iii) Comparing equations (ii) and (iii),

2 = 2

0

q

2 ma

= 20

q

2 ma

….(iv)

Hence, time period,

T = 2

= 2

02 ma2

q

….[using (iv)]

508. K.E. = W = Fx = qEx 509. From charge configuration, at the centre,

electric field is non-zero. Potential at the

centre due to 2q charge 2q

2qV

r

and potential due to – q charge

q

qV

r (r = distance of centre point)

Total potential 2q q qV V V V 0

510. Potential at centre O of the square

O

0

QV 4

4 (a / 2)

Work done in shifting (– Q) charge from centre to infinity

O OW Q(V V ) QV 2

0

4 2 Q

4 a

2

0

2Q

a

511 Potential at the centre of square,

9 6

49 10 50 10V 4 90 2 10 V

2 / 2

Work done in bringing a charge (q = 50 C) from to centre (O) of the square is

0 0W q(V V ) qV

6 4W 50 10 90 2 10 64J

512. Electrical potential, V = 0

Q

4 R

Electric field E = 0 inside a conductor 514. At centre E = 0 and V = 0

515. Potential at the centre is 5 0

1 q

4

l

The electric field due to the oppositely placed charges cancel and net electric field is

20

1 q

4

l

.

516. We know that potential at surface of a sphere

is V = K q

r= 10 V (Given) ….(i)

where, q and r is the charge and radius of the small drop respectively.

As the volume of 27 small drops and that of 1 large drop must be same, hence

3427 r

3

= 34R

3

where, R is radius of large drop. R = 3r Now the total charge on large drop is

Q = 27q Hence, potential at surface of this drop is

V = KQ

R=

(27q)K

(3r)= 9

Kq

r

V = 90 V ….[Using eq (i)] 517. V = 6x – 8xy – 8y + 6yz

Ex = V

x

= (6 – 8y) = 2

Ey = V

y

= (8x 8 + 6z) = 10

Ez = V

z

= 6y = 6

– q – q

2q

r r

r

E– q E2q E– q

O

C

B A

D

+Q

E

E E

E

– Q

– Q + Q

103

Chapter 01 : Electrostatics

E = 2 2 2x y zE E E = 4 100 36 = 140

= 2 35 N/C

F = qE = 2 2 35 = 4 35 N

518. E

= dV

dr

r

= x i

+ y j

+ z k

E

= V V V

i j zx y z

At point (1, 1, 0);

E

= 6 i

5 j

2 k

= (6 i

+5 j

+ 2 k

)

519.

2A

A A

2B BB

1mVE W (q)V2

1E W (4q)VmV2

A

B

V 1

V 2

520. From conservation of energy, P.E = K.E. Ui Uf = (K.E.)f (K.E.)i

1 2

0 i f

q q 1 1

4 r r

= 2f

1mv

2 ….{ (K.E)i = 0}

1 22f

0

q q2v

m 4

i f

1 1

r r

= 6 3 9

3

2 1 10 2 10 9 10 1 1

1 10 1 10

2fv 32400

vf = 180 m/s

521. AB

= B A

= ˆ ˆa 4j 3k

Work done = F AB

= 0

ˆ ˆ ˆq k a 4j 3k2

0

3q a

2

….{ k j 0

; k k 1

}

522. In the given condition angle between p

and E

is zero. Hence potential energy U pEcos0 pE min.

Also in uniform electric field Enet = 0.

523. C k; V 1

k; U

1

k 524. High k means good insulating property and

high x means able to withstand electric field gradient to a higher value.

525. k

Cd

;

i. Glass: 4 40

200.2 2

;

ii. Quartz: 3 30

7.50.4 4

;

iii. Teflon: 2

21 ;

iv. Mica: 5 50

60.8 8

526. E = 0k

= k0E

= 2.2 8.85 1012 3 104 6 107 C/m2

527. If nothing is said, it is considered that battery is disconnected. Hence charge remains the same.

Also air0

V d

and medium0

tV d t

k

m

a

td t

V kV d

m

68 6

V 6120 8

Vm = 45 V 528. 0

0

AkC 4 r

d

r = Radius of sphere of equivalent capacity

4

3

Ak 100 10 6r 4.77 m

4 d 1 10 4 3.14

529. Electric field inside parallel plate capacitor

having charge Q at place where dielectric is

absent = 0

Q

A and where dielectric is present

= 0

Q

kA

530. 1

2

C

C =

1

3

In series, charge on each capacitor is same.

1

2

V

V = 1

2

q / C

q / C = 2

1

C

C =

3

1

Dividing 10 V in the ratio 3 : 1, we get V1 = 7.5 V, V2 = 2.5 V 531. On connecting O at A, 4 F capacitor is

charged to a constant potential (E). As connection of O is switched over to B, the

total charge on 4 F capacitor is shared

between 4 F and 2 F capacitor is 4

4 2 =

2

3

of original charge. 532. CP = C1 + C2

S

1

C =

1

1

C +

2

1

C = 1 2

1 2

C C

C C

104 

Physics Vol‐II (Med. and Engg.) 

104

CS = 1 2

1 2

C C

C C

According to given condition CP – CS = 3 F

C1 + C2 – 1 2

1 2

C C

C C = 3

As C1 = C2 = C

2C – 2C

2 C = 3

4C2 – C2 = 6 C 3 C2 = 6 C C = 2 F

533. For total capacitance 1

C =

1

C + 1

C C +

1

C

= 1

4 +

1

8 +

1

4= 2 1 2

8

= 5

8

C = 8/5 F and V = 10 V Then energy

= 1

2CV2 =

1

2

8

5 10–6 (10)2

= 8 10–5 J. 534. 21

U CV2

Now if V is constant, then U is greatest when Ceq is maximum. This is when all the three are in parallel.

535. The charge q remains unchanged as the battery is disconnected. The capacitance C decreases if the separation between the plates is increased. Now, energy stored U = q2/2C. Since q remains the same and C is decreased, U will increase.

536. In steady state condition, no current flows through line (1). Hence total current

I 2.5

1A(1 1 0.5)

Potential difference across line (2) = potential

difference across capacitor = 1 2 = 2 volt So, charge on capacitor = 5 2 = 10 C 537. Volume of 8 drops will be same as volume of

1 large drop formed by combining smaller drops.

3 34 48 r R

3 3

R = 2r the capacitance of bigger drop is C = 40R = 40 (2r) = 2C 538. In steady state potential difference across

capacitor = 2 V. So charge on capacitor Q = 10 2 = 20 C 539. In steady state potential difference across

capacitor V2 = potential difference across resistance

22

1 2

RR V

R R

Hence V2 depends upon R2 and R1. 540. Cnet = 5 F Qnet = 5 8 = 40 C We know, Q2F = 2 8 = 16C Q4F = Q12F = Qnet Q2F ...{9F|| 3F = 12F}

= 40 16 = 24C Voltage across 4F and 12F can be given as, V4F + V12F = V

V4F = V 12 F

12 F

Q

C

= 8

24

12 = 6V

V12F = 2V i.e. V9F = 2V Q9 F = 9 2 = 18 C Q = Q4 F + Q9 F = 42 C

E = 9 6

2

kQ 9 10 42 10

r 30 30

= 420 N/C

541. Capacitance will be increased when a dielectric is introduced in the capacitor but potential difference will remain the same because battery is still connected. So according to

q = CV, charge will increase i.e., Q > Q0 and

U = 0 0 0 0

1 1QV ,U Q V

2 2

Q > Q0 so U > U0 542. According to energy conservation, energy

remains the same

parallel seriesU U 2 21 1 C(nC)V V

2 2 n

V = nV (V = potential difference across series

combination)

1

2

1

5 F

Line (2)

Line (1)

2.5V

+ –

V1

V2

V

R1

R2

R3 C

105

Chapter 01 : Electrostatics

543. Energy stored in the capacitor 21CV 100

2

6 3 2 61

10 10 ×(100 10 ) 100 5 10 J2

Electric energy costs = 108 paise per kWH

6

108 paise

3.6 10 J

Total cost of charging6

6

5 10 108

3.6 10

= 150 paise

544. 1 2eq 3

1 2

C C 2 6C C 4 5.5 F

C C 2 6

Energy supplied 2 6(E) qV CV 22 10 J

P.E.stored 2eq

1(U) C V

2

= 6 21

5.5 10 (2)2

U = 611 10 J

Energy lost 6E U 11 10 J

545. Power =

21CV

2t

=6 2

3

1 40 10 (3000)

2 2 10

= 90 kW 546. For the soap bubble, Pin Pout = Pexcess = PST Pelectro

= 2

20

4T q

r 2A

=

2

2 20

4T q

r 2(4 r )

= 2

2 40

4T q

r 32 r

For equilibrium, Pin = Pout

4T

r=

2

2 40

q

32 r

q = 2 4

016 2 4 r T

r

=

2 4016 8 r T

r

q = 4r2 08 T

r

547. Let the charge of each drop be q

C = q

V

q = CV charge of final drop, Q = nq Radius of each small drop is r and big drop is R volume remains same,

34R

3 = 34

n r3

R3 = nr3 R = (n)1/3 r Potential on big drop

V = KQ

R

V = 1/3

Knq

n r

Ratio of energy stored in big drop to small drop,

1/3

Knq1 QQVU QV n r21 KqU qVqV q2 r

1/3

U nq Knqr

U n rq Kq

U

U

=

5/3n

1 548. Let 2F

be the force exerted by charge q2 on +q

Let 1F

be the force exerted by charge q1 on +q

Fnet = F2 F1

=2 2

2 2

kq kq

(L x) (L x)

= 22 2

1 1kq

(L x) (L x)

= 2 2

22 2

(L x) (L x)kq

(L x) (L x)

= 2 2 2 2

22 2 2

L x 2Lx L x 2Lxkq

(L x )

= 22 2 2

4Lxkq

(L x )

= 24

4Lxkq

L

….{ x << L}

= 23

4xkq

L

Fnet = 2

3

4kq x

L

Fnet = ma

2

3

4kq x

L = ma ….(i)

a = 2 x (for S.H.M) ….(ii)

2 x = 2

3

4kq x

mL ….from (i) and (ii)

2 = 2

3

4kq

mL

=2

3

4kq

mL

T = 2

T = 23

2

mL

4kq

T 3L T (L)3/2

2L

q1 = +q+q

q2 = +q2F

1F

x Displaced position of particle

m

106 

Physics Vol‐II (Med. and Engg.) 

106

1. A charge of 5.8 1018 C (A) does exists. (B) does not exist. (C) exists for a short time. (D) exists only if placed in higher orbits. 2. Two point charges repel each other with a

force of 100 N. One of the charges is increased by 10% and other is reduced by 10%. The new force of repulsion at the same distance would be

(A) 100 N (B) 121 N (C) 99 N (D) 90 N 3. Three charges – q1, + q2 and – q3 are placed as

shown in figure. The x component of the force on – q1 is proportional to

(A) 2 32 2

q qsin

b a (B) 2 3

2 2

q qcos

b a

(C) 2 32 2

q qsin

b a (D) 2 3

2 2

q qcos

b a

4. For corona discharge, the conductor should

have a (A) spherical end (B) flat end (C) pointed end (D) concave end 5. A square is centered at origin having sides

parallel to x and y-axis. It has surface charge density (x, y) = 0 xy within its boundaries. The measure of charge of each side of square is Q. What is the total charge on the square?

(A) 40Q2 (B) 20a

3 (C) 0a

2 (D) Zero 6. Two positive charges of 20 coulomb and Q

coulomb are situated at a distance of 60 cm. The neutral point between them is at a distance of 20 cm from the 20 coulomb charge. Charge Q is

(A) 30 C (B) 40 C (C) 60 C (D) 80 C 7. An electric dipole, when held at 30 with

respect to a uniform electric field of 104 NC–1 experiences a torque of 9 1026 Nm. Calculate dipole moment of the dipole.

(A) 1.7 10–29 Cm (B) 1.8 10–30 Cm (C) 1.8 10–29 Cm (D) 1.7 10–30 Cm

8. A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will

(A) increase four times. (B) be reduced to half. (C) remain the same. (D) be doubled. 9. Two charged conducting spheres of radii R1

and R2 separated by a large distance are connected by a long wire. The ratio of the charges on them is

(A) 1

2

R

R (B) 2

1

R

R

(C) 2122

R

R (D)

2221

R

R

10. The electric potential due to the nucleus of the hydrogen atom at a distance of 5.3 10–11 m is 27.2 V. What is the potential due to the helium nucleus at the same distance?

(A) 27. 2 V (B) 54.4 V (C) 13.6 V (D) 20.4 V 11. The electric field in a region is radially

outward with magnitude E = Ar. Find the charge contained in a sphere of radius 20 cm. Given: A = 100 V m–2

(A) 9.89 10–11 C (B) 7.85 10–11 C (C) 8.89 10–11 C (D) 8.19 10–11 C 12. Two positive point charges 12 C and 8 C

are placed 10 cm apart in air. How much work must be done to bring them close by 6 cm?

(A) 2.8 J (B) 3.8 J (C) 4.8 J (D) 13.0 J 13. The radius of a spherical conductor is 0.9 m.

Its capacity is (A) 1 pF (B) 10 pF (C) 100 pF (D) 1000 pF 14. The capacitance of the earth viewed as a

spherical conductor of radius 6408 km is (A) 600 F (B) 712 F (C) 980 F (D) 1424 F 15. When two capacitors are connected in series,

the equivalent capacitance is 15

4F. When

they are connected in parallel the equivalent capacitance is 16 F. The individual capacitance are

(A) 5 F, 11 F (B) 6 F, 10 F (C) 4 F, 12 F (D) 8 F, 8 F

–q3

–q1 +q2

y

a

bx

Topic Test

107

Chapter 01 : Electrostatics

16. The capacities of two conductors are C1 and C2 and their respective potentials are V1 and V2. If they are connected by a thin wire then the loss of energy will be

(A) 1 2 1 2

1 2

C C (V V )

2(C C )

(B) 1 2 1 2

1 2

C C (V V )

2(C C )

(C) 2

1 2 1 2

1 2

C C (V V )

2(C C )

(D) 1 2 1 2

1 2

(C C ) (V V )

C C

17. If a point charge q0 is placed at the centre of a

thin wire ring of radius r, then what is the increased value of force exerted on stretching the wire? [Given, q is the electric charge on the wire.]

(A) 02 2

0

qq

4 r (B) 0

2 20

qq

2 r

(C) 02 2

0

qq

r (D) 0

2 20

qq

8 r

1. (B) 2. (C) 3. (C) 4. (C) 5. (D) 6. (D) 7. (C) 8. (C) 9. (A) 10. (B) 11. (C) 12. (D) 13. (C) 14. (B) 15. (B) 16. (C) 17. (D)

Answers to Topic Test