chapter16.pptx

T . Norah Ali Almoneef 1

CHAPTER 16 ELECTRIC FORCES, FIELDS ,AND

POTENTIALLightning

Becomes very “negative”

Becomes very “positive”


Arbitrary numbers of protons (+) and electrons (-) on a comb and in hair (A) before and (B) after combing. Combing transfers electrons from the hair to the comb by friction, resulting in a negative charge on the comb and a positive charge on the hair.

• If a positively charged rod is brought near a trickle of water, the water moves towards it. What happens if we use a negatively charged rod?

2


Electric Charge• Types:

– Positive• Glass rubbed with silk • Missing electrons

– Negative• Rubber/Plastic rubbed with fur• Extra electrons

• Arbitrary choice – convention attributed to ?

• Units: amount of charge is measured in [Coulombs]

• Empirical Observations:– Like charges repel– Unlike charges attract

4T . Norah Ali Almoneef

CONSERVATION OF ELECTRIC CHARGE

• In the process of rubbing two solid objects together, electrical charges are NOT created. Instead, both objects contain both positive and negative charges. During the rubbing process, the negative charge is transferred from one object to the other and this leaves one object with an excess of positive charge and the other with an excess of negative charge. The quantity of excess charge on each object is exactly the same.


Charge PropertiesCONSERVATION OF ELECTRIC CHARGE

– Charge is not created or destroyed, only transferred.• , however, it can be transferred from one object to another. The net amount

of electric charge produced in any process is zero.• Quantization

– The smallest unit of charge is that on an electron or proton. (e = 1.6 x 10-19 C)• It is impossible to have less charge than this• It is possible to have integer multiples of this charge

Q Ne

– A coulomb is the charge resulting from the transfer of 6.24 x 1018 of the charge carried by an electron.

– The magnitude of an electrical charge (q) is dependent upon how many electrons (n) have been moved to it or away from it.Mathematically,

• Electric Charge and Electrical Forces:

• Electrons have a negative electrical charge.

• Protons have a positive electrical charge.

• These charges interact to create an electrical force.– Like charges produce repulsive forces – so they repel each other (e.g.

electron and electron or proton and proton repel each other).

– Unlike charges produce attractive forces – so they attract each other (e.g. electron and proton attract each other).


It is impossible to have less charge than thisIt is possible to have integer multiples of this charge

Q NeHow many electrons constitute 1 mC?

19#

1.6 10totalq

electronsx

7

Electric Force - Coulomb’s Law• Consider two electric charges: q1 and q2

• The electric force F between these two charges separated by a distance r is given by Coulomb’s Law

• The constant k is called Coulomb’s constant and is given by

221

r

qkqF

T . Norah Ali Almoneef

229 /CNm109 k• Coulomb law

– The electrical force between two charged bodies is directly proportional to the charge on each body and inversely proportional to the square of the distance between them.

8

• The coulomb constant is also written as

• 0 is the “electric permittivity of vacuum”– A fundamental constant of nature

According to the superposition principle the resultant force on a point charge q equals the vector sum of the forces exerted by the other point charges Qi that are present:

2

212

00 Nm

C 1085.8 where

4

1

k

221

04

1

r

qqF

T . Norah Ali Almoneef

• The force between two charges gets stronger as the charges move closer together.

• The force also gets stronger if the amount of charge becomes larger.


the resultant force on any one of them equals the vector sum of the forces exerted by the various individual charges. For example, if four chargesare present, then the resultant force exerted by particles 2, 3, and 4 on particle 1 is

1 22e e

q qF k

r1 2

12 122ˆ

e

q qk

rF r

or


example

• A positive charge of 6.0 x 10 -6C is 0.030m from a second positive charge of 3.0 x 10 -6C. Calculate the force between the charges.

= (8.99 x 109 N m2/C2 ) (6.0 x 10 -6C) (3.0 x 10 -6C)

( 0.030m )2

= (8.99 x 109 N m2/C2 ) (18.0 x 10 -12C)

(9.0 x 10 -4 m2)

= + 1.8 x 10 -8 N

1 22e e

q qF k

r


Three point charges, q1 = - 4 nC, q2 = 5 nC, and q3 = 3 nC, are placed as in the Fig.


Coulomb's Law• The force between charges is directly

proportional to the magnitude, or amount, of each charge.

• Doubling one charge doubles the force. • Doubling both charges quadruples the force.

• The force between charges is inversely proportional to the square of the distance between them.

• Doubling the distance reduces the force by a factor of 22 = (4), decreasing the force to one-fourth its original value (1/4).

• .


• Compare electrical to gravitational force in a hydrogen atom

NF

NF

gravity

electrical

47210

312711

8210

2199

1061.31053.0

1011.91067.11067.6

102.81053.0

106.1109

•Electron and proton attract each other 1040 times stronger electrically than gravitationally


An object, A, with +8.25 x 10-6 C charge, has two other charges nearby. Object B, -3.5 x 10-6 C, is 0.030 m to the right. Object C, +2.50 * 10-6 C, is 0.050 m below. What is the net force and the angle on A?V

X

Y


Example• What is the magnitude of the electric force of attraction between

an iron nucleus (q=+26e) and its innermost electron if the distance between them is 1.5 x 10-12 m

.The magnitude of the Coulomb force isF = kq1q2/r2

= (9.0 X 109 N · m2/C2)(26)(1.60 X 10–19 C)(1.60 X10–19 C)/(1.5 X10–12 m)2 = 2.7 X 10–3 N.


Example• We have a charge of 5 C at X = 2 m.• What will the force on a charge of -4 C be at X = 0 m.

• F = kq1 q2 / r2

• , F = 5 C x (4C) x 9x109 / 4 m2

• F = 45 x109 N


Given that q = -13 µC and d = 13 cm, find the direction and magnitude of the net electrostatic force exerted on the point charge q1 in Figure. F12(force on q1 from q2)= k (q1)(q2)/(d)2

= (9x109 N m2/C2) (13x10 -6C)(26x10-6C)/(0.13m)2

= 180 Ntowards q2 (attractive)

F13 (force on q1 from q3)= k(q2)(q3)/(2d)2

= (9E9 N m2/C2) (13x10-6C)(-39x10-6C)/(0.26m)2

= +67.4 Naway from q3 (repulsive)

Net force = (180 – 67.4)=112 N Towards q2

q1=+q q2=-2q Q3=+3q

example


Two point charges lie on the x axis. A charge of -9.5 µC is at the origin, and a charge of + 2.5 µC is at x = 10.0 cm.

(a) What is the net electric field at x = -2.0 cm?Electric field from charge –9.5x10-6C isE1 = (9x10+9 Nm2/C2)(9.5x10-6C)/(0-(-0.02m))2 E1 = 2.14E8 N/C field points towards charge –9.5x10-6 (positive x-direction)Electric field at x=-0.02 m from charge 2.5x-6 C isE2 = (9x10+9 Nm2/C2)(2.5x10-6C)/(0.10m-(-0.02m))2 E2 = 1.56E6 N/Cfield points away from charge 2.5x10-6 (negative x-direction)Net electric field E = 2.14E8 i N/C - 1.56E6 i N/C= [2.12e+08] i N/C

example


Three point charges, q1 = -1.2 x 10-8 C, q2 = -2.6 x 10-8 C and q3 = +3.4 x 10-8 C, are held at the positions shown in the figure, where a = 0.16 m

example


exampleTwo electrostatic point charges of +20.0 μC and –30.0 μC exert attractive forces on each other of –145 N .What is the distance between the two charges?

mr

xxxxr

F

kr

x

qq

193.0145

309 10102010669

21


Three point charges lie along the x axis as shown in Figure. The positive charge q1 ! 15.0 *C is at x ! 2.00 m,the positive charge q2 ! 6.00 *C is at the origin, and the resultant force acting on q3 is zero. What is the x coordinate of q3?

Solution Because q3 is negative and q1 and q2 are positive,

For the resultant force on q3 to be zero, F23 must be equal in magnitude and opposite in direction to F13. Setting the magnitudes of the two forces equal,

T . Norah Ali Almoneef 21T . Norah Ali Almoneef 21

x1 x2

Consider two charges located on the x axisThe charges are described byq1 = 0.15 C x1 = 0.0 m

q2 = 0.35 C x2 = 0.40 m

Where do we need to put a third charge for that charge to be at an equilibrium point?

At the equilibrium point, the forces from the two charges will cancel.

third charge to be at an equilibrium point when

21 FF

22

21

232

231

)4.0(

)4.0(

x

q

x

q

x

qkq

x

qkq

x= 0.12m or =

0.72m X


ExampleSuppose two charges having equal but opposite charge are separated by6.4 × 10-8 m. If the magnitude of the electric force between the chargesis 5.62 ×10–14 N, what is the value of q


exampleCharged spheres A and B are fixed in position, as shown, and have charges of +7.9 x 10-6 C and -2.3 x 10-6 C, respectively. Calculate the net force on sphere C, whose charge is +5.8 x 10-6 C.

])1015(

103.2

)1025(

109.7[108.5109

22

6

22

669

cF

LeftthetoNF ..26.1


9.71.122

F

T . Norah Ali Almoneef 25 25

Example:What is the force between two charges of 1 C separated by 1 meter?

221

r

qkqF F= 9

2

9

1091

11109x

xxx

Example:

What is the electric force between a +5 mC charge and a –3 mC charge separated by 3 cm?

F = (9 x 109 Nm2/C2)(.005C)(.003C) / (.03 m)2

= 150,000,000 N


How far apart would two objects, each with a charge of 1Coulomb, have to be so that they only exerted a 1 Newton electric force on one another?


Object A has a charge of +2 μC, and object B has a charge of +6 μC. Which statement is true about the electric forces on the objects?

• FAB = –3FBA

• FAB = –FBA

• 3FAB = –FBA

• FAB = 3FBA

• FAB = FBA

• 3FAB = FBA

From Newton's third law, the electric force exerted by object B on object A is equal in magnitude to the force exerted by object A on object B and in the opposite direction.


Where do I have to place the + charge in order for the force to balance, in the figure at right?

• Force is attractive toward both negative charges, hence could balance.

• Need a coordinate system, so choose total distance as L, and position of + charge from -q charge as x.

• Force is sum of the two force vectors, and has to be zero, so

• A lot of things cancel, including Q, so our answer does not depend on knowing the + charge value. We end up with

• Solving for x, , so slightly less than half-way between.A

-q

-2q

x

L

0)(

22221

x

qQk

xL

qQkFFF

22

1

)(

2

xxL

22

)(2

2

x

xL

x

xL

LL

x 412.021


A charged particle, with charge Q, produces anelectric field in the region of space around itA small test charge, qo, placed in the field, willexperience a force

EF q0

Electric Field•Mathematically,•Use this for the magnitude of the field•The electric field is a vector quantity The direction of the field is defined to be the direction of the electric force that would be exerted on a small positive test charge placed at that point

qF

E

0

•For a point charge

16.2The Electric Field


16.2The Electric Field• The force (Fe) acting on a “test” charge (qo) placed in an electric

field (E) isFe = qoE

• Note the similarity of the electric force law to Newton’s 2nd Law (F=ma)

• Formal definition of electric field:– the electric force per unit charge that acts on a test charge

at a point in space or E = Fe/qo

•The electric field exists whether or not there is a test charge present•The Superposition Principle can be applied to the electric field if a group of charges is present


Electric Field• The electric force on a positive test charge q0 at a distance r from a single charge q:

rr

qqkFe ˆ

20

• The electric field at a distance r from a single charge q:

rr

qk

q

FE e ˆ

20

EqFe

0


A positive charge is released from rest in a region of electric field. The charge moves:

a) towards a region of smaller electric potential

b) along a path of constant electric potential

c) towards a region of greater electric potential

Example 9

A positive charge placed in an electric field will experience a force given by EqF

Therefore

Since q is positive, the force F points in the direction opposite to increasing potential or in the direction of decreasing potential

EqF


+

-+

+

+

-

-

The electric field at a given point Pis the sum of the electric fields due to every point charge

1

21

ˆ

N

ii

Ni

ii

i

E E

qrk

r

P

A charge distribution


THE ELECTRIC FIELD

F am

The electric force on a charge q is

F Eq

which, together with Newton’s 2nd Law,can be used to calculate the motion of an electric charge, of mass m

Newton’s 2nd Law for an electric charge can be written as

q

mE a If E is constant, both in direction and

magnitude, so to is the acceleration ofthe charge.

Note that the acceleration depends on the charge to mass ratio.


Electric field & linePoint charge•The lines radiate equally in all directions•For a positive source charge, the lines will radiate outward

For a negative source charge, the lines will point inward


– A map of the electrical field can be made by bringing a positive test charge into an electrical field.

• No two field lines can cross. • You can draw vector arrows to indicate the direction of the

electrical field.This is represented by drawing lines of force or electrical field lines,

• These lines are closer together when the field is stronger and farther apart when it is weaker.

•Field lines must begin on positive charges (or from infinity) and end on negative charges (or at infinity). The test charge is positive by convention

•the magnitude of the electric force will increase proportionally with an increase in charge and/or and increase in the electric field magnitude

• The line must be perpendicular to the surface of the charge

•The number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the magnitude of the charge


Shark


Fish to detect object


Electric Field lines


example

15363 1087.8)8.9()106.0(3

4)1000(

3

4 grmg

eN

q

mgqE

120

109.1462/1087.8 1715


Which of the following statements about electric field lines associated with electric charges is false?

• 1) Electric field lines can be either straight or curved.

• 2) Electric field lines can form closed loops.

• 3 )Electric field lines begin on positive charges and end on negative charges.

• 4 ) Electric field lines can never intersect with one another.

example


(1) 4 E0

(2) 2 E0

(3) E0

(4) 1/2 E0

(5) 1/4 E0

You are sitting a certain distance from a point charge, and you measure an electric field of E0. If the charge is doubled and

your distance from the charge is also doubled, what is the electric field strength now?

Remember that the electric field is: E = kQ/r2. Doubling

the charge puts a factor of 2 in the numerator, but

doubling the distance puts a factor of 4 in the

denominator, because it is distance squared!! Overall,

that gives us a factor of 1/2.


+2 +1+1 +1d d

1)

2)

3) the same for both

Between the red and the blue charge, which of them experiences the greater electric field due to the green charge? +2

+1

Both charges feel the same electric field due

to the green charge because they are at the

same point in space!

2r

QkE


+2 +1+1 +1d d

1)

2)

3) the same for both

Between the red and the blue charge, which of them experiences the greater electric force due to the green charge?

+2

+1

The electric field is the same for both charges, but the

force on a given charge also depends on the

magnitude of that specific charge.

qEF


exampleFind electric field at point P in the figure

CNr

kqE /18

4

108109 99

2

o87.306

2.1

6.1tan 1


ExampleIn figure shown, locate the point at which the electric field is zero? Assume a = 50cm

E1 = E2

d = 30cm

ExampleFind the electric field at point p in figure .due to the charges shown.



E1 = E2

d = 30cm




E1 = E2

d = 30cm


Ex = E1 - E2 = -36´104N/CEy = E3 = 28.8´104N/C Ep = [(36´104)2+(28.8´104)2 ] = 46.1N/C

q = 141o


Example

0tot E

30°

E1E2

E3

Three identical charges (q = –5.0 mC) lie along a circle of radius 2.0 m at angles of 30°, 150°, and 270°, as shown. What is the resultant electric field at the center of the circle?

0)21

21

()30sin30sin(

)(

030cos30cos

22220

20

1

321

221

02

01

r

qk

r

qk

r

qk

r

qkEEE

EEEEr

qkEE

EEE

y

yyy

x


Example A +100 C point charge is separated from a -50 C charge by a distance of 0.50 m as shown below. (A) First calculate the electric field at midway between the two charges. (B) Find the force on an electron that is placed at this point and then calculate the acceleration when it is released.

+

Q1

_Q2

E

In part A we found that E = 2.1x107 N/C and is directed to the right.

q

FE

EeEqF

CNCF 719 101.2106.1 N12104.3

amF

( to left )

m

Fa 2

1831

12

107.3101.9

104.3s

mkg

N

( to left )


the Field on Electric Dipole

y

ar

kqE

2)(

The total field at P is

yarar

kqEE

}

)(

1

)(

1{

22

y

ar

karE ]

)(

4[

222

When r is much greater than a we can neglect a in the denominator then

ykqa

Er

3

4 rE 3

1

16.3 Electric Field due to arrangements of charges

yar

kqE

2)(


Note• •the dipole electric field reduces as 1/r3, instead of 1/r of a single charge.• although we only calculate the fields along z-axis, it turns out that this also applies to all direction.• p is the basic property of an electric dipole, but not q or d. Only the product qd is important.

E

E

p

xE

xEEE

xExE

3

22

22

4

11

;

R

kqaaR

aRaRkq

aR

kq

aR

kq

A set of two (equal and opposite) charges separated by a distance


Example Find the electric field due to electric dipole shown in figure along x-axis at point p which is a distance r from the origin. then assume r>>a

Solution

When x>>a then


A test charge of +3 μC is at a point P where an external electric field is directed to the right and has a

magnitude of 4 × 106 N/C. If the test charge is replaced with another test charge of –3 μC, the external electric field at P

• A )is unaffected

• B )reverses direction

• C )changes in a way that cannot be determined

There is no effect on the electric field if we assume that the source charge producing the field is not disturbed by our actions. Remember that the electric field is created by source charge(s) (unseen in this case), not the test charge(s).


A Styrofoam ball covered with a conducting paint has a mass of 5.0 × 10-3 kg and has a charge of 4.0 μC. What electric field directed upward will produce an electric force on the ball that will balance the weight of the ball?(a) 8.2 × 102 N/C(b) 1.2 × 104 N/C(c) 2.0 × 10-2 N/C(d) 5.1 × 106 N/C

The magnitude of the upward electrical force must equal the weight of the ball. That is:


•An electric dipole consists of two equal and opposite charges•The high density of lines between the charges indicates the strong electric field in this region

Two equal but like point charges•At a great distance from the charges, the field would be approximately that of asingle charge of 2q•The bulging out of the field lines between the charges indicates the repulsion between the charges•The low field lines between the charges indicates a weak field in this region


Since the 4 C charge feels a force, there must

be an electric field present, with magnitude:

E = F / q = 12 N / 4 C = 3 N/C

Once the 4 C charge is replaced with a 6 C

charge, this new charge will feel a force of:

F = q E = (6 C)(3 N/C) = 18 N

Q

1) 12 N

2) 8 N

3) 24 N

4) no force

5) 18 N

In a uniform electric field in empty space, a 4 C charge is placed and it feels an electrical force of 12 N. If this charge is removed and a 6 C charge is placed at that point instead, what force will it feel?


Determine the point (other than infinity) at which the total electric field is zero

we will call the position of the negative charge x = 0,which means the positive charge is at x = 1 m. We will call the position where electric field is zero x. The distance from this point tothe negative charge is just x, and the distance to the positive charge is 1 + x. Now write down the electric field due to each charge:

example

39.0 ,82.1

65.2

)106()105.2(

)106(

)105.2(

22

2

6

2

6

2

6

2

6

)1(

)1(

)1(

x

xkxk

xk

xk

xx

xx

xE

xE

neg

posWe wrote down the distance x the distance to the left of the negative charge. A negative value of x is then inthe wrong direction, in between the two charges, which we already ruled out. The positive root, x = 1.82, means a distance 1.82mto the left of the negative charge. This is what we want.


A charge q1 = 7.0 µC is located at the origin, and a second charge q2 =5.0 µC is located on the x axis, 0.30 m from the origin .Find the electric field at the point P,which has coordinates (0, 0.40) m.

T . Norah Ali Almoneef 61T .Norah Ali Almoneef 61


Electric fields of Concentric spherical shells

RE

KQ2If the smaller shell has a radius R : (Out side)

Surface area of the sphere is : RA2

4

AQK

E4


Planar Symmetry

Two conducting plates with charge density 1

All charges on the two faces of the plates For two oppositely charged plates placed near each other, E field

outer side of the plates is zero while inner side the E-field= 2 1/0

For two oppositely charged plates placed near each other, E field outer side of the plates is zero while inner side the E-field = E = 2π KQ / A


E field of a single uniformly charged plate• Charged infinite plane: E = 2pks (s: Q/A)• E = 2π KQ / A

• E =4π KQ / A

For two oppositely charged plates placed near each other, E field outer side of the plates is zero while inner side the E-field

0

0inside

outsideE

QE

A


16.4 ELECTRIC POTENTIAL• The electrostatic force is a conservative (=“path independent”)

force• It is possible to define an electrical potential energy function with

this force• Work done by a conservative force is equal to the negative of the

change in potential energy

• There is a uniform field between the two plates• As the positive charge moves from A to B, work

is done

• WAB=F d=q E d

• ΔPE =-W AB=-q E d– only for a uniform field


SI Unit of Electric Potential: joule/coulomb=volt (V)

Potential Difference (=“Voltage Drop”)

• The potential difference between points A and B is defined as the change in the potential energy (final value minus initial value) of a charge q moved from A to B divided by the size of the charge• ΔV = VB – VA = ΔPE /q

• Potential difference is not the same as potential energy

The electric potential V at a given point is the electric potential energy U of a small test charge q0 situated at that point divided by the charge itself:

If we set at infinity as our reference potential energy,

The electric potential difference between any two points i and f in an electric field.


• Another way to relate the energy and the potential difference: ΔPE = q ΔV

• Both electric potential energy and potential difference are scalar quantities

• A special case occurs when there is a uniform electric field• VB – VA= -Ed

• Gives more information about units: N/C = V/m•The electric potential energy U and the electric potential V are not the same. The electric potential energy is associated with a test charge, while electric potential is the property of the electric field and does not depend on the test charge.

• It is equal to the difference in potential energy per unit charge between the two points.

• the negative work done by the electric field on a unite charge as that particle moves in from point i to point f.

• A larger charge would involve a larger amount of PEe, but the ratio of that energy to the charge is the same as it would be if a smaller charge was in that same place.


Energy and Charge Movements• A positive charge gains electrical potential energy when it is

moved in a direction opposite the electric field• If a charge is released in the electric field, it experiences a force

and accelerates, gaining kinetic energy– As it gains kinetic energy, it loses an equal amount of electrical potential

energy• A negative charge loses electrical potential energy when it moves

in the direction opposite the electric field

• When the electric field is directed downward, point B is at a lower potential than point A

• A positive test charge that moves from A to B loses electric potential energy

• It will gain the same amount of kinetic energy as it loses potential energy


• Electric field always points from higher electric potential to lower electric potential.

• A positive charge accelerates from a region of higher electric potential energy (or higher potential) toward a region of lower electric potential energy (or lower potential).

– it moves in the direction of the field, Its electrical potential energy decreases, Its kinetic energy increases

• A negative charge accelerates from a region of lower potential toward a region of higher potential.

– It moves opposite to the direction of the field– Its electrical potential energy decreases– Its kinetic energy increases


• If a charged particle moves perpendicular to electric field lines, no work is done.

• If the work done by the electric field is zero, then the electric potential must be constant

• Thus equipotential surfaces and lines must always be perpendicular to the electric field lines.

if d E

General Considerations

V We

q0 V is constant


Electrical Potential Energy in a Uniform Electric Field• If a charge is released in a uniform electric field at a constant

velocity there is a change in the electrical potential energy associated with the charge’s new position in the field.

• PEe = -qE d The unit: Joules• The negative sign indicates that the electrical potential energy will

increase if the charge is negative and decrease if the charge is positive.

• The V in a uniform field varies with the displacement from a reference point.

• V = E d• The displacement is moved in the direction of the field.• Any displacement perpendicular to the field does not change the electrical

potential energy.


POTENTIAL ENERGY IN A UNIFORM FIELD

The Electric Field points in the direction of a positive test charge.

+ Charge - Charge

Along E Loses PEe Gains PEe

Opposite E Gains PEe Loses PEe


Example: An electron accelerates from rest through an electric potential of 2 V. What is the final speed of the electron?

Answer: The electron acquires 2 eV kinetic energy. We have

and since the mass of the electron is me = 9.1 × 10–31 kg, the speed is

, J 10 3.2eV

J 10 1.6 eV] 2[

2

1 19192

vme

. m/s 10 4.8kg 10 9.1

J 10 3.22 531–

19

v


exampleAn electron (mass m = 9.11×10-31kg) is accelerated in the uniform field E (E = 1.33×104 N/C) between two parallel charged plates. The separation of the plates is 1.25 cm. The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate, as seen in the figure. With what speed does it leave the hole?

F = qE = ma

a = qE/m

Vf 2 = vi2 + 2a(d)

Vf 2 = 2ad = 2(qE/m)d

• Vf 2 = 2ad = 2(qE/m)d

= 2 (1.9 x 10 -19C) (1.33×104 N/C) (1.25m) / 9.11×10-31kg= 8.3 x 10 6 m/s

m = 9.11×10-31kg E = 1.33×104 N/C d = 1.25 cm

GIVEN:


“Electric field lines always point in the direction of decreasing electric potential”

The unit: V m-1 Or NC-1


e-


But a more useful concept is the electric potential energy of each charge

e-

e-

e-

e-

The greater the magnitude of the charge, the greater is the electric potential energy


Clicker Question • In the figure, a proton moves from

point i to point f in a uniform electric field directed as shown. Does the electric field do positive, negative or no work on the proton?

A: positive B: negative C: no work is done on the proton


Electric Potential Energy with a Pair of Charges

• A single point charge produces a non-uniform electric field.

• PEe = kq1q2/r• The reference point for PEe is assumed to be at infinity. • The ground is usually the reference point for PEg.

• The PEe is positive for like charges and negative for unlike charges.


Work and Electrical Potential Energy• In order to bring two like charges near each other work must be

done. (W = Fd)• In order to separate two opposite charge, work must be done.• Remember: whenever work gets done, energy changes form.

The potential energy will change to kinetic energy.


• Since the electrical potential energy can change depending on the amount of charge you are moving, it is helpful to describe the electrical potential energy per unit charge.

• Electric potential: the electrical potential energy associated with a charged particle divided by the charge of the particle.

• A larger charge would involve a larger amount of PEe, but the ratio of that energy to the charge is the same as it would be if a smaller charge was in that same place.


16.4Electrical Potential Energy of Two Charges• V1 is the electric potential due to q1 at some

point P1

• The work required to bring q2 from infinity to P1 without acceleration is q2E1d=q2V1

• This work is equal to the potential energy of the two particle system

r

qqkVqPE 21

e12

• If the charges have the same sign, PE is positive– Positive work must be done to force the two charges near one

another– The like charges would repel

• If the charges have opposite signs, PE is negative– The force would be attractive– Work must be done to hold back the unlike charges from

accelerating as they are brought close together


Potential Difference Near a Point Charge• An electric potential exists at some point in an electric field

regardless of whether there is a charge at that point.• The electric potential at a point depends on only two quantities:

the charge responsible for the electric potential and the distance (r)from this charge to the point in question.

• V = kcq/r

• Voltage is a way of using numbers (quantitative) to describe an electric field.

• Electric fields are measured in volts over a distance. This means the larger the E the larger the V.

• When an E is attracting or repelling an object, we instead could say that the object is being driven by the voltage.


Calculate the electric potential, V, at 10 cm from a -60mC charge.

Equation: Answer: V = -5.4x 106V

Calculate the electric potential, V, at the midpoint between a 250 mC charge and a -450 mC separated by a distance of 60 cm.

Equation: Answer: V = -6.0x106V

Example:


If you want to move in a region of electric field without changing your electric potential energy. You would move

a) Parallel to the electric fieldb) Perpendicular to the electric field

Example


General Points for either positive or negative charges

The Potential increases if you move in the direction opposite to the electric field

andThe Potential decreases if you move in the same direction as the electric field

Electric Potential

Electric Potential is a scalar field

it is defined everywhere

but it does not have any direction

it doesn’t depend on a charge being there


What is the potential difference between points A and B?ΔVAB = VB - VA

a) ΔVAB > 0 b) ΔVAB = 0 c) ΔVAB < 0

E

A

BC

ExamplePoints A, B, and C lie in a uniform electric field.

Since points A and B are in the same relative horizontal location in the electric field there is on potential difference between them

The electric field, E, points in the direction of decreasing potential


E

A

BC

Point C is at a higher potential than point A.

True False


As stated previously the electric field points in the direction of decreasing potential

Since point C is further to the right in the electric field and the electric field is pointing to the right, point C is at a lower potential

The statement is therefore false


Compare the potential differences between points A and C

and points B and C.

a) VAC > VBC b) VAC = VBC c) VAC < VBC

E

A

BC


In Example 4 we showed that the the potential at points A and B were the same

Therefore the potential difference between A and C and the potential difference between points B and C are the same

Also remember that potential and potential energy are scalars and directions do not come into play


If a negative charge is moved from point A to point B, its electric potential energy

a) Increases. b) decreases. c) doesn’t change.

E

A

BC


The potential energy of a charge at a location in an electric field is given by the product of the charge and the potential at the location

As shown in Example 4, the potential at points A and B are the same

Therefore the electric potential energy also doesn’t change


The Volt• The commonly encountered unit joules/coulomb is called the volt, abbreviated V, after the

Italian physicist Alessandro Volta (1745 - 1827)

• With this definition of the volt, we can express the units of the electric field as

• For the remainder of our studies, we will use the unit V/m for the electric field.

1 V = 1 J

1 C

m

V

C

J/m

C

N

][

][][

qF

E

The Joule• The joule is a measure of work accomplished on an object.• It is also a measure of potential energy or how much work an object can do.• In the English system the unit of work and energy is the ft x lb.• F = m x a For a falling object a = g, so F = m x g• Energy is force x distance.• E = F x d• For a falling object d=h (h=height)• E = F x h• PE= m x g x h


The Electron Volt

• The electron volt (eV) is defined as the energy that an electron (or proton) gains when accelerated through a potential difference of 1 V

• 1 V=1 J/C 1 eV = 1.6 x 10-19 J

There is an additional unit that is used for energy in addition to that of joules

A particle having the charge of e (1.6 x 10-19 C) that is moved through a potential difference of 1 Volt has an increase in energy that is given by

eV

joulesqVW

1

106.1 19


Potential energy

Calculate the work done bringing a 250 mC charge and a -450 mC from infinity to a distance of 60 cm apart.

Equation: = 9x109(250x10-6)(-450x10-6)

60x10-2

PE = -1.7x103J


Example• A proton is placed between two parallel

conducting plates in a vacuum as shown.The potential difference between the two plates is 450 V. The proton is released from rest close to the positive plate.

• What is the kinetic energy of the proton when it reaches the negative plate?

+ -

= V(+)-V(-)The potential difference between the two plates is 450 V.

The change in potential energy of the proton is DU, and DV = DU / q (by definition of V), soDU = q DV = e[V(-)-V(+)] = -450 eV


ExampleSuppose an electron is released from rest in a uniform electric field whose magnitude is 5.90 x 103 V/m. (a) Through what potential difference will it have passed after moving 1.00 cm? (b) How fast will the electron be moving after it has traveled 1.00 cm? (a) |DV| = Ed = (5.90 x 103 V/m)(0.0100 m) =

59.0 V

(b) q |DV| = mv2/2 v = 4.55x106 m/s

ExampleAn ion accelerated through a potential difference of 115 V experiences an increase in kinetic energy of 7.37 x 10 –17 J. Calculate the charge on the ion.

qV= 7.37x10-17 J , V=115 V q = 6.41x10-19 C


Example

A particle has a mass of 1.8x10-5kg and a charge of +3.0x10-5C. It is released frompoint A and accelerates horizontally until it reaches point B. The only force actingon the particle is the electric force, and the electric potential at A is 25V greater thanat C. (a) What is the speed of the particle at point B? (b) If the same particle had anegative charge and were released from point B, what would be its speed at A?

AABB EPEmvEPEmv 2212

21

BAAB EPEEPEmvmv 2212

21

BAoAB VVqmvmv 2212

21


ExampleThe work done by the electric force as the test charge (+2.0x10-6C)

moves from A to B is +5.0x10-5J.(a) Find the difference in EPE between these points.(b) Determine the potential difference between these points.

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPEJ100.5EPEEPE 5 ABAB W

V 25C102.0

J100.56-

5

o

ABAB q

WVV

BAABW EPEEPE (a)

J100.5EPEEPE 5 ABAB W

(b) V 25C102.0

J100.56-

5

o

ABAB q

WVV


Example 3: A proton is moved from the negative plate to the positive plate of a parallel-plate arrangement. The plates are 1.5cm apart, and the electric field is uniform with a magnitude of 1500N/C.

a) How much work would be required to move a proton from the negative to the positive plate?

b) What is the potential difference between the plates?

c) If the proton is released from rest at the positive plate, what speed will it have just before it hits the negative plate?

cosxFW qEFE

JW

mC

NCW

xqEW

18

19

106.3

)015)(.1500)(106.1(

1


Example:

Two 40 gram masses each with a charge of -6µC are 20cm apart. If the two charges are released, how fast will they be moving when they are a very, very long way apart. (infinity)

s

m

rm

kqv

rm

kqv

mvr

qk

mvmvr

qqk

KKU

EWE

E

f

0.2)4)(.2(.

)106)(109(

2

1

2

1

2692

22

22

2221

21

0


Problem:

An electron is released from rest in an electric field of 2000N/C. How fast will the electron be moving after traveling 30cm?

_v = 0m/s

_v = ?

30cm

31

19

1011.9

)3)(.2000)(106.1(2

v

s

mv 71045.1

KUE

2

21

mvqEd

mqEd

v2


• F = qE q=+0.045·10-6 C > 0, Force parallel to E Charge accelerates to left

Work done on charge = qEd = Change in Kinetic Energy W = (0.045·10-6C)(1200 V/m)(0.05m) = 54. ·10-3V ·C =0.054J (Kf-Ki) = W (1/2) mvf

2 – 0 = 0.054 J vf

2 = 2 (0.054 J) / ( 3.5 ·10-3kg)= 30.8 m2/s2

vf =5.6 m/s

F=qE

Ignore triangle on sketch.


Example: An electron accelerates from rest through an electric potential of 2 V. What is the final speed of the electron?

The electron acquires 2 eV kinetic energy. We have

and since the mass of the electron is me = 9.1 × 10–31 kg, the speed is

, J 10 3.2eV

J 10 1.6 eV] 2[

2

1 19192

vme

. m/s 10 4.8kg 10 9.1

J 10 3.22 531–

19

v


An a- particle, mass = 6.7x10-27 kg, initially at rest travels 25 cm through a uniform electric field of 250 N/C. Calculate the

potential difference across the 25 cm path

Equation: V =Ed Answer: 62.5 V

What is the a-particle’s speed after 25 cm of travel?

Equation: qEd = ½ mv2 Answer:7.7x104m/s


How much energy (work) is necessary to bring three point charges from infinity to the vertices of the right triangle shown below.

2.0 mC3.0 mC

-4.0mC

5.0 cm

3.0 cm

4.0 cm

Equation:PE = kq1q2

Answer: PE = -2.16JPE = 1.8 JPE = -1.8 JPE = -2.16J

r


Calculate the work done bringing a 250 mC charge and a -450 mC from infinity to a distance of 60 cm apart.

= 9x109(250x10-6)(-450x10-6)60x10-2

PE = -1.7x103Jrqq

KPE 21

example :The potential difference between to charge plates is 500V. Find the velocity of a proton if it is accelerated from rest from one plate to the other.

500V

+

+

-

-+

sm

v

v

mqV

v

mvqV

KUE

5

27

19

2

101.3

1067.1)500)(106.1(2

221

High

Potential

Low Potential

Positive charges move from high to low potentialNegative charges move from low to high potential

example


Which of the following statements is false? A. The total work required to assemble a collection of discrete charges is the

electrostatic potential energy of the system. B. The potential energy of a pair of positively charged bodies is positive. C. The potential energy of a pair of oppositely charged bodies is positive. D. The potential energy of a pair of oppositely charged bodies is negative. E. The potential energy of a pair of negatively charged bodies is negative.

The figure depicts a uniform electric field. Along which

direction is the increase in the electric potential a maximum?

example

example


Potential Difference in a Uniform field

E

+Q +Q

+Q

A B

C0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd||QEdU AC

d||

||EdVAC


The potential at a point due to a unit positive point charge is found to be V. If the distance between the charge and the point

is tripled, the potential becomes

A. V/3. B. 3V. C. V/9. D. 9V. E. 1/V 2 .

example

The figure shows two plates A and B. Plate A has a potential of 0 V and plate B a potential of 100 V. The dotted lines represent equipotential lines of 25, 50, and 75

V. A positive test charge of 1.6 × 10–19 C at point x is transferred to point z. The

electric potential energy gained or lost by the test charge is

A. 8 × 10–18 J, gained. B. 8 × 10–18 J, lost. C. 24 × 10–18 J, gained. D. 24 × 10–8 J, lost. E. 40 × 10–8 J, gained.

example


Two parallel metal plates 5.0 cm apart have a potential difference between them of 75 V. The electric force on a positive charge of 3.2 × 10–19 C at a point midway between the plates is approximately

A. 4.8 × 10–18 N. B. 2.4 × 10–17 N. C. 1.6 × 10–18 N. D. 4.8 × 10–16 N.E. 9.6 × 10–17 N.

example

When +2.0 C of charge moves at constant speed from a point with zero potential to a point with potential +6.0 V, the amount of work done is

A. 2 J. B. 3 J. C. 6 J. D. 12 J. E. 24 J.

example


The concept of “potential difference" or “voltage" in electricity is similar to the concept of "height" in gravity, or “pressure” in fluids


A positive charge of 6 nC attracts a negative charge of 5 nC from a distance of 4 mm. What is the work done? What does the sign mean?

A point charge of +3 μC is located at the origin of a coordinate system and a second point charge of -6 μC is at x = 1.0 m. At what point on the x-axis is the electrical potential zero?

1. -0.25 m2. +0.25 m3. +0.33 m4. +0.75 m


Example :

Compute the energy necessary to bring together the charges in the configuration shown below:

Calculate the electric potential energy between each pair of charges and add them together.

JU

U

r

qqkU

72.02.

)104)(104()109(

12

669

12

2112

JU

JJJU

total

total

72.

)72.()72.(72.


Two point charges of values +3.4 and +6.6 μC respectively, are separated by 0.20 m. What is the potential energy of this 2-charge system?

1. +0.34 J2. -0.75 J3. +1.0 J4. -3.4 J

What will be the electrical potential at a distance of 0.15 m from a point charge of 6.0 μC?

1. 5.4 x 104 V2. 3.6 x 105 V3. 2.4 x 106 V 4. 1.2 x 107 V

T . Norah Ali Almoneef 116T .Norah Ali Almoneef 116

Example1. The potential difference between the two terminals on a battery

is 9 volts. How much work (energy) is required to transfer 10 coulombs of charge across the terminals?

V = 9.0 V W= ? q= 10 c 9.0 = W/10

W = 90 J

2. The work required to transfer 30 coulombs of charge across two terminals is 50 joules.

What is the potential difference? V= ? W= 50 J q= 30 c

V=50/30

V =1.7 V

Example


Problem

Show that the amount of work required to assemble four identical point charges of magnitude Q at the corners of a square of side s is 5.41keQ2/s.

1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 5.41

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s


Example

Consider a positive and a negative charge, freely moving in a uniform electric field. True or false?

(a) Positive charge moves to points with lower potential.(b) Negative charge moves to points with lower potential.(c) Positive charge moves to a lower potential energy

position.(d) Negative charge moves to a lower potential energy

position

–Q +Q 0

+V

–V

(a) True(b) False(c) True(d) True


PROBLEM• A proton is released from rest in a uniform electric field with

a magnitude of 8 E 4 V/m. The proton is displaced 0.5 m as a result. A) Find the potential difference between the proton’s initial and final positions. B) Find the change in electrical potential energy of the proton as a result of this displacement.

E = 8 E 4 V/m q = + 1.6 10 -19 C d = 0.5 m

A) V = -Ed V = - (8 E 4 V/m)(0.5 m) = V = - 40000 V

B) V = PEe/q PEe = V q (-40000 V)(+1.6 10 -19 C) = - 6.4 E-15 J


The Potential due to a Point Charge

1 1( ) ( )AB

B A

V V B V A kqr r

Although only changes in potential are physicallyrelevant, it is often convenient to choose the locationof the zero of the potential. Fora car battery, this is typically the car’schassis; for an electrical outlet itis the ground. For an isolated pointcharge, it is convenient to choose the potential to be zero at infinity


The Potential of a Point Charge

1 1( ) ( )AB

B A

V V B V A kqr r

The potential difference between two points A and B from a point charge

can be re-written as1 1

( ) ( )B A

V B V A kqr r

When rA = infinity the last term vanishes. We are free to choose V(A) as we please, e.g., V(A) = 0.With this choice, the potential of a point charge becomes ( )

kqV r

r


• Superposition principle applies• The total electric potential at some

point P due to several point charges is the algebraic sum of the electric potentials due to the individual charges– The algebraic sum is used because

potentials are scalar quantities

Electric Potential of Many Point Charges• Electric potential is a SCALAR not a vector.• Just calculate the potential due to each individual

point charge, and add together! (Make sure you get the SIGNS correct!)

q1

q5

q4

q3

q2

i i

i

r

qkV

r1

r2

r3

r4

r5

P


Potential from more than one charge

Principle of superposition

...4

1

2

2

1

1

0 r

Q

r

QV

r

QV

04

1

...21 VVV

1

1

01 4

1

r

QV

Total Potential is sum of all individual potentials

Each potential is

Thus Total potential is

Which can be written


Electric Potential of a single charge

+

r if V = 0 at rA=

r

QV

1

4 0

E

B

A

It can be shown that2

0

1

4 r

QE

Remember that

rEV so

This looks a bit like the formulae for the potential in a Uniform Field

||EdVAC Potential energy Arbitrary shape

Potential differenceArbitrary shape


Potential Energy in 3 charges

r

QV

04

1

Q2

Q1

Q3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU

Energy when we bring in Q2

Now bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU

So finally we find


Example:

Pictures from Serway & Beichner

How many electrons should be removed from an initially uncharged spherical conductor of radius 0.300 m to produce a potential of 7.5 kV at the surface?

Substituting given values into V = = 7.50 x 103 V

=

r

qke

)m300.0(

)/CNm10x99.8( 229 q

C 1050.2 7q

N = 1.56 x 1012 electrons


Example 1: The electron in the Bohr model of the atom can exist at only certain orbits. The smallest has a radius of .0529nm, and the next level has a radius of .212m.

a) What is the potential difference between the two levels?

b) Which level has a higher potential?

+er1

r2

r

qkV

11 r

ekV

VV 2.27100529.

106.1)109(

9

199

1

VV 79.6100212.

106.1)109(

9

199

2

VVdiffpotential 4.2079.62.27

r1 is at a higher potential.


Example: Finding the Electric Potential at Point P (apply V=keq/r).

5.0 mC-2.0 mC

V1060.3)m0.4()m0.3(

)C100.2()C/Nm1099.8(

,V1012.1m0.4

C100.5)C/Nm1099.8(

3

22

6229

2

46

2291

V

V

Superposition: Vp=V1+V2

Vp=1.12104 V+(-3.60103 V)=7.6103 V


PROBLEMS• As a particle moves 10 m along an electric field of strength 75 N/C,

its electrical potential energy decreases by 4.8 x10 – 16 J. What is the particle’s charge?

• What is the potential difference between the initial and final locations of the particle in the problem above?

• An electron moves 4.5 m in the direction of an electric field of strength 325 N/C/ Determine the change in electrical potential energy.

1) E = 75 N/C PEe = - 4.8 x10 -16 J d = 10 m q = ? Must be a + q since PEe was lost

PEe = - qEd q = (- 4.8 x10 -16)/(-(75)(10)) =Q = +6.4 E -19 C2) V = PEe/q (-4.8 x10 -16 J)/(6.4 E -19 C)

V = - 750 V3) q = - 1.6 x10 -19 C d = 4.5 m E = 325 N/C PEe = ? PEe = - qEd -(-1.6 x10 -19 C)(325 N/C)(4.5 m) = 2.3 x10 -16 J


Example: Consider three point charges q1 = q2 = 2.0 C and q3 = -3 C which are

placed as shown. Calculate the net force on q1 and q3


The force on q1 is F1 = F12 + F13.


Similarly, F3 = F31 + F32


V 240

m 60.0

C100.8CmN1099.8

m 20.0

C100.8CmN1099.8 82298229

AV

V 0

m 40.0

C100.8CmN1099.8

m 40.0

C100.8CmN1099.8 82298229

BV

Example

At locations A and B, find the total electric potential.


Example A charge +q is at the origin. A charge –2q is at x = 2.00 m on the x axis. For what finite value(s) of x is (a) the electric field zero ? (b) the electric potential zero ?

x2 + 4.00x – 4.00 = 0 (x+4.83)(x0.83)=0\x = - 4.83 m

(other root is not physically valid)

0)00.2(

2

x

q

x

qV

x = 0.667 m and x= -2.00 m

0)00.2(

222

x

q

x

qkE


EXAMPLE: POTENTIAL OF GROUP OF POINT CHARGES

• In the drawing on the right, q = 2.0 μC and d = 0.96 m. Find the total potential at the location P, assuming that the potential is zero at infinity.

V 4.9

296.0

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV


The Potential due to a Point Charge:


The three charges in Fig. , with q1 = 8 nC, q2 = 2 nC, and q3 = - 4 nC, are separated by distances r2 = 3 cm and r3 = 4 cm. How much work is required to move q1 to infinity?

Example


Calculate the electric potential, V, at 10 cm from a -60mC charge.

Equation:

Answer: V = -5.4x 106V

Calculate the electric potential, V, at the midpoint between a 250 mC charge and a -450 mC separated by a distance of 60 cm.

Equation: Answer: V = -6.0x106V

Example

Example


If a negative charge is moved from point A to point B, its electric potential energy

a) Increases. b) decreases. c) doesn’t change.

E

A

BC

Example

Points A, B, and C lie in a uniform electric field.

The potential energy of a charge at a location in an electric field is given by the product of the charge and the potential at the location

As shown in Example 4, the potential at points A and B are the same

Therefore the electric potential energy also doesn’t change

144


E. Potential & Potential Energy vsElectric Field & Coulomb Force

If we know the potential field this allows us to calculate changes in potential energy for any charge introduced

Coulomb Force is thus Electric Field multiplied by charge

Electric Field is Coulomb Force divided by test charge

D Potential is D Energy divided by test charge

D Energy is D Potential multiplied by test charge

0Q

UV

0Q

FE

VQU EF Q


Electric Potential Difference

The electric potential energy depends on the charge present

We can define and electric potential V which does not depend on charge by using a “test” charge

EdQU 0

Change in potential is change in potential energy for a test charge divided by the unit charge

EdQ

UV

0

Remember that for uniform field

0Q

UV


Parallel-Plate Capacitor• The capacitor consists of two parallel plates• Each have area A• They are separated by a distance d• The plates carry equal and opposite charges• When connected to the battery, charge is pulled off one

plate and transferred to the other plate• The transfer stops when DVcap = DVbattery

147


CapacitorsStoring a charge between the

plates• Electrons on the left plate are

attracted toward the positive terminal of the voltage source

• This leaves an excess of positively charged holes

• The electrons are pushed toward the right plate

• Excess electrons leave a negative charge

+ -

+_

+ _


Capacitance of parallel plates

+Q -Q

The bigger the plates the more surface area over which the capacitor can store charge C A

Moving plates togeth`er Initially E is constant (no charges moving) thus V=Ed decreases charges flows from battery to increase V C 1/d

Never Ready+

E

V

Capacitance – is a measure of the capacitor’s ability to store electric energy

V

qC

where: q – magnitude of the charge

V – potential/potential difference

C - capacitance


In SI units,

q – coulomb (C)

V – volt (V)

C – coulomb/volt = A Farad is very large Often will see µF or pFfarad (F)


• The capacitance of a device depends on the geometric arrangement of the conductors

• For a parallel-plate capacitor whose plates are separated by air:

d

AC o

150

(C does not depend on Q or V)

[V = Ed, E=Q/(A e0), V = Qd / (A e0)]

For any geometry, capacitance scales as Area/distance


Electric Field• A uniform electric field (The field strength is

the same magnitude and direction at all points in the field)can be produced in the space between two parallel metal plates.

• The plates are connected to a battery.

E

The field strength at any point in this field is:

E = field strength (Vm-1)V = potential difference (V)d = plate separation (m)

d

VE


Capacitance

The constant of proportionality C is the capacitance which is a property of the conductor

VQ VCQ V

QC

Experiments show that the charge in a capacitor is proportional to the electric potential difference (voltage) between the plates.

d

A

V

QC

dA

QEdV

0

0

To increase C, one either increases , increases A, or decreases d.

Note that if we doubled the voltage, we would not do anything to the capacitance. Instead, we would double the charge stored on the capacitor.

However, if we try to overfill the capacitor by placing too much voltage across it, the positive and negative plates will attract each other so strongly that they will spark across the gap and destroy the capacitor. Thus capacitors have a maximum voltage!


Example

The plates of the capacitor are separated bya distance of 0.032 m, and the potential differencebetween them is VB-VA=-64V. Between thetwo equipotential surfaces shown in color, thereis a potential difference of -3.0V. Find the spacingbetween the two colored surfaces.

mV100.2m 0.032

V 64 3

d

VE

m105.1mV100.2

V 0.3 33

E

Vd


The two plates of a capacitor hold +5000mC and -5000mC, respectively, when the potential difference is 200V. What is the capacitance?

Equation:

How much charge flows from a 12 V battery when connected to a 20 mF capacitor?

Equation: q = CV

240 mC

FCVV

qC

25200

105000 6

q =20F x12V =


Example• How strong is the electric field between the plates of a 0.80

mF air gap capacitor if they are 2.0 mm apart and each has a charge of 72 mC?

m / V 45000002.0

90

d

VE

V 908.0

72

C

QV

Find the capacitance of a 4.0 cm diameter if the plates are separated by 0.25 mm.

Example

FC 10 115.410 325.0

10 326.11085.8 12

128.85 10 F/m rACd

2 2 3 2π 0.02 m 1.26 10 mA r

FC 10 115.410 325.0

10 326.11085.8 12


ExampleWhen a potential difference of 150 V is applied to the plates of a parallel-plate capacitor, the plates carry a surface charge density of 30.0 nC/cm2. What is the spacing between the plates?

0 AQ V

d

12 2 2

09 2 4 2 2

8.85 10 C N m 150 V4.42 m

30.0 10 C cm 1.00 10 cm m

Vd

What is the charge on a 250 microfarad capacitor if it has been charged to 12 V?

q = CV = (250E-6 F)(12 V) = 3E-3 Coulombs

Example


Circular parallel plate capacitor

r r

s

r = 10 cm

A = r2 = (.1)2

A = .03 m 2

S = 1 mm = .001 m

S

AC

0

FC 10103

pFC 300

001.

03.)10( 11C Farad

Volt

Coulomb

p = pico = 10-12

Show Demo Model, calculate its capacitance , and show how to charge it up with a battery.


Example: Thundercloud• Suppose a thundercloud with horizontal dimensions of 2.0 km by 3.0

km hovers over a flat area, at an altitude of 500 m and carries a charge of 160 C.

• Question 1:– What is the potential difference between

the cloud and the ground?• Question 2:

– Knowing that lightning strikes requireelectric field strengths of approximately2.5 MV/m, are these conditions sufficientfor a lightning strike?


Example:• Question 1• We can approximate the cloud-ground system as a parallel plate

capacitor whose capacitance is

• The charge carried by the cloud is 160 C, which means that the “plate surface” facing the earth has a charge of 80 C

• 720 million volts

C 0A

d

(8.85·10-12 F/m)(2000 m)(3000 m)

500 m0.11 F

V q

C

80 C

0.11 F7.2108 V

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

…++++++++++++ …


• Question 2• We know the potential difference between the cloud and

ground so we can calculate the electric field

• E is lower than 2.5 MV/m, so no lightning cloud to ground– May have lightning to radio tower or tree….

V/m 10644.1500

1082.7

d

VE


A parallel-plate capacitor has an area of 5.00 cm2 and the plates are separated by a distance of 2.50 mm Calculate the capacitance.

What is the capacitance if it has .014 V across it when it has a charge of 2.13x10-15 C?

C = q/V = (2.13E-15 C)/(.014 V) = 1.52x10-13 F

pFC

C

d

AC

8.1105.2

10585.8

3

412

0

Problem

Problem


• If a spark jumps across a 1mm gap when you reach for a doorknob, what was the potential difference between you and the knob and how much charge was on you?

• Assume your finger and the knob form a parallel plate capacitor with area 1 cm2 and breakdown electric field = 3MV/m.

V = Q/C V = EdQ = CV=CEd = (A 0/d)(E d) = A 0 EQ = (0.01m)2 (8.8510-12 F/m)(3.06 V/m) Q = 2.7 10-15 (C/V)(Vm2/m 2) = 2.7 femto-Coulomb

Problem


Problem(a) What plate area is required if an air-filled, parallel plate capacitor

with a plate separation of 2.6 mm is to have a capacitance of 12 pF?

(a) C = A 0/d,A = Cd/ 0 = (12·10 -12 F)(2.6 ·10 -3 m) / (8.8510 -12 F/m)A = 3.5 ·10-3 m2 = (6cm)2

A 0.25 μ F capacitor is connected to a 400 V battery. What is the charge on the capacitor?1. 1.2 x 10-12 C2. 1.0 x 10-4 C 3. 0.040 C4. 0.020 C

Problem


The two plates of a capacitor hold +5000mC and -5000mC, respectively, when the potential difference is 200V. What is the capacitance?

Equation: Answer 25 mF

How much charge flows from a 12 V battery when connected to a 20 mF capacitor?

Equation: q = CV Answer: 240 mC

V

qC

Problem

Problem

chapter16.pptx

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