asdasdads2014 Electric Field Assignment Guide

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Anglo-Chinese Junior College Assignment SolutionsTopic : Electric Field H2 (9646)

JC1 2014 Page 1 of 8

1Afor a uniform field magnitude of the electric field strength is E = -

d

V hence

at X, Eis stronger; electric field lines points from higher to lower potential2 (a) C

1. Diagram shows both charges must be positive as E field lines point awayfrom the charges.

2. At null pt,p q

p q

Q Q

r r

2 24 4

Henceq q

p p

r Q

r Q

2

2, therefore Qq> Qp

3. Electric potential due to a point charge is given by V =r4 0

Q. Resultant

potential at X is > 0.

(b)

Note: Equipotential lines always perpendicular to E-field lines

3C Use

QqF

r

24

andQq

Ur

4

4 Dwork done by an external agent is equal to the increase in electrical potentialenergy

Wext= U = V= q(VfVi) Wext= -q(V2-V1) = q(V1V2)

5 B noteelectric field strength is a vector while electric potential is a scalar6 (a) Dwork done by an external agent Wext= V =q(VfVi) = (-1.6x10

-19)(100 -400)

(b)Note: 1.strength of field is stronger at the

sharp point and field lines should beradial there2.Equipotential lines areperpendicular to E field lines


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7 D Since charge is positive, positive work is done when change in potential ispositive. Direction of field lines is from higher to lower potential.C Decrease in potential hence ve work done on positive charge.A and Cequipotential line hence no work done.

8 (a) the electric field strength at a point in an electric field is the electric force per unitcharge on a stationarypositive test point chargeplaced at that point.It acts in the direction that the positive test charge would then move.

(b) Since electric field strength E = F/q,the force acting on the charge +q is F = +q E in the direction of the field since it

ispositive charge.Work done W = Fd = + qEd

(c) the potential difference V between two points is the work done per unit chargewhen a small positive test point charge is moved between the two points.V =W/q = + qEd/q

= Ed

9 Bbetween two parallel oppositely charge plates the numerically value to theelectric

potential gradient is constant i.e.d

V = constant

40

V=

20

XYV VXY =

V40

20

10 Bfor Y to be suspended stationary the electric force on Y must be upwards to

balance its weight. This can only be possible if the top plate X and the sphere Yare of opposite polarity while the lower plate Z and the sphere Y are of the samepolarity.

11 (a)2

4 rQE

o

(Note this the formula for point charges, not valid for parallel

plates, which is as given above)

Ex=1.6 x 10-19x 9 x 109/ (4.1 x -10) 2= 8.57 x 109NC-1

EY=17.14 x 109NC-1

Deducingthat triangle is a right angle triangle (do not assume unless checked)It can shown easily that angle ZXY = XYZ = 45o

Hence the resultant field E can be deduced usingE2= Ex

2+ Ey2

E =1.92 x 1010NC-1

(b) (i) Magnitude of Force = 4.8 x 10-

x 1.92 x 10 = 9.22 x 10-

N(ii) Sketch giving direction or by calculation, Angle = 63.4 0to Ex

Direction of force is opposite to the direction of E-field since negative charge

(c) 1 coulomb is a very large charge (and it would totally alter the electric field which

is being created) (See Tutorial Q1)

63.4o Ex

EY


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(d)

(e) Note that an obvious pt where the resultant potential is zero is when they areequidistant apart which is the tip of the equipotential line given(This point is themid-pt of the line joining the two positive charges.)

Hence the full equipotential line can be drawn using the fact that equipotentiallines must be drawn perpendicular to the field lines using this point as areference.

12 (a) The electr ic potent ial ,Vat a point is defined as the work done per unitchargeby an external agent in bringing a small positive point test chargefrom infinity to that point, without producing any acceleration of the charge.

The electr ic p otent ial energy, Uof a system of charges is defined as the workdone by an external agent to bring the point charges from infinity to theirrespective positions in the system, without producing any acceleration of the

charges.

(b) (i)

200 9000

8800

f iV V V

V

Hence gain in electrical potential is 8800 V

(ii)

19

15

1.6 10 8800

1.41 10

U q V

J

Hence the loss in electrical potential energy is 1.4110-15J.

-3

+1 +2

Equipotential line of zero

potential (ans to (f))


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12 (iii) Assuming no energy losses,Applying the principle of conservation of energy,

15

1.41 10

ki i kf f

kf ki f i

k

E U E U

E E U UE U

J

Hence the gain in kinetic energy is 1.41 10-15J(iv)

2 15

6

10 1.41 10

2

5.56 10

-1

m s

kf ki k E E E

mv

v

(c) When negatively charged particles move from a point of lower electric potential to a pointof higher electric potential, they gain kinetic energy but loses electric potential energy.

13 D

14 (a) (i)E =

d

V

=)100.5(

)03000(3

x

= 6.0x105Vm-1 or NC-1

14 (a) (ii) To determine time consider the horizontal motionHorizontal displacement, Sx=uxt40x10-3 =(3.9x106) t

t = 1.026 x 10-8 s= 1.0 x 10-8 s

(b) (i)1F = qE and F = ma

a = 5100.6m

qxE

q

m

14 (b) (i)2

Consider the vertical motion of the charged particlevy = uy+ ayt

vy = 0 +5

100.6m

qxE

q

m

=6.153 x10-3q

m

=6.2 x10-3q

m


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14 (ii)0

x

y3.4tan

v

v = 0.07519

6.153 x10-3 6q /(3.9 x10 ) .m

0 07519

710x4.766m

q

= 4.8x107 C kg-1

15 (a)

x/cm VA/106V VB/10

6V VR/10

6V EA/ 10

6N C

-1 EB/ 10

6N C

-1 ER/10

6NC

-1

1.4 1.57 0.058 1.63 112 0.55

4.0 0.548 0.076 0.624 13.7 0.96

6.0 0.366 0.102 0.468 6.09 1.70 8.0 0.274 0.153 0.427 3.43 3.82 0.39

10.0 0.219 0.306 0.525 2.19 15.3 13.1

11.4 0.192 1.020 1.21 1.69 170 168


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16 (a) (i) A gravitational field of force is a region in space where a mass will experience agravitational force.

(ii) An electric field of force is a region in space where an electric charge willexperience an electric force.

(b) Gravitational force acting on a mass is always in the same direction as that ofthe gravitational field.

The electric force acting on a electric charge is in the same direction as that ofthe electric field if the charge is positive but in the opposite direction if the chargeis negative.

(c)The electric force between 2 electrons at a distance r apart is FE= 2

0

2

r4

e

The gravitational force between 2 electrons at a distance r apart is FG= 2

2

e

r

mG

Hence

2

e0

2

G

E

mG

1

4

e

F

F

= 4.158 x 1042= 4.16 x 1042

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