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SF027 1

UNIT 1:Electrostatics

The study of electriccharges at rest, theforces between themand the electric fields

associated with them.



SF027 2

1.1 COULOMB’S LAW

LEARNING OUTCOMES:

a) State Coulomb’s Law

b) Sketch the force diagram and apply Coulomb’s Law for

a system of point charges

224 r

kQq

r

QqF

o



SF027 3

There are two kinds of charges innature – positive and negative

charge.

Like charges repel.

Unlike charges attract.

The magnitude of the force, Fbetween two point charges is

given by Coulomb’s Law.



Coulomb’s Law

• States – the magnitude of the electrostatic(Coulomb/electric) force between two pointcharges is proportional to the product of the

charges and inversely proportional to thesquare of the distance between them.

+ +r

2q1q

F

F



SF027 5

Mathematically,

chargespointobetween twdistance:r

2

21

r

qkqF

2-29 C N m10 x0.9k constant(Coulomb)ticelectrosta:

2

21

r qqF

where

force(Coulomb)ticelectrostaof magnitude:F chargeof magnitude:, 21 qq

+ +

r

2q1qF

F

Coulomb’s law



SF027 6

• Since

04

1k

, hence the Coulomb’s law can be written as

2

21

0 r

qq

4

1F

where

air)or(vacuumspacefreeof typermittivi:0

).( 21212

0 m N C 10 x858

+

r -

2q1q F

F



• If q1 and q2 are charges of opposite sign, theforce (F ) acting on each charge is attractive asshown in figure below.

– This mean that F is directed towards theneighbouring charge and will result in bothcharges moving towards each other.

• If q1 and q2 are both positive or both negativecharges, the force (F ) acting on each charge isrepulsive.

– This mean that F is directed away from theneighbouring charge and will result in aseparation of the two charges if they arefree to move.



SF027 8

• The S.I. unit of charge is coulomb (C ).

• Note :

– The sign of the charge can be ignored when substitutinginto the Coulomb’s law equation.

– The sign of the charges is important in distinguishing thedirection of the electric force .



Example 1 :Two point charges, q1=-20 nC and q2=90 nC,

are separated by a distance of 4.0 cm asshown in figure below.

Find the magnitude and direction of

a. the electric force that q1 exerts on q2.

b. the electric force that q2 exerts on q1.

(Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2)

-

cm04 .

+ 2q1q



SF027 10

Solution: q1=2.0 x 10-8 C, q2=9.0 x 10-8 C, r=4.0 x 10-2 m

21F 12 chargeonchargebyforce:

-

cm04 .

+ 2q1q 12F 21F

where

12F 21 chargeonchargebyforce:



SF027 11

a. By applying the Coulomb’s law equation :

b. By using the Coulomb’s law equation :

Conclusion :

– The magnitude of both forces is the same but opposite in direction

– obey the Newton’s third law.

– The characteristic of electric force exert on both charges isattractive force.

22

889

12

10 x4

10 x0910 x0210 x09F

)(

).)(.)(.(

2112 F F

2

2112

r

qkqF

N 10 x01F 2

12

.

Direction : to the left (q1)

2

12

21 r

qkq

F

N 10 x01F 2

21

.

Direction : to the right (q2)



Example 2 :Three point charges lie along the x-axis as

shown in figure below.

Calculate the magnitude and direction of the totalelectric force exerted on q2.

(Given Coulomb’s constant, k = 9.0 x 10

9

N m

2

C-2)

C 4q2 C 2q1 -+ +

C 6 q3

cm03 . cm05 .



SF027 13

Solution: r 12=3.0 x 10-2 m, r 23=5.0 x 10-2 m

By applying the Coulomb’s law equation :

32F

12F C 4q2 C 2q1 -+ +

C 6 q3

cm03 . cm05 .

2

12

2112

r

qkqF

N 10 x08F 13

12 .




SF027 14

And

Therefore, the total force exerted on q2 is given by

223

32

32 r

qkq

F

N 10 x6 8F 13

32 .


32122 F F F

N 10 x6 16 F 13

2 .

32122 F F F

1313

2 10 x6 810 x08F ..




Example 3 :

Figure below shows the three point charges are

placed in the shape of triangular.

Determine the magnitude and direction of the

resultant electric force exerted on q1. Given q1=-1.2 C, q2=+3.7 C, q3=-2.3 C, r12=15 cm,r13=10 cm, =32 and k = 9.0 x 109 N m2 C-2.

2q1q +

3q

-

-

12r 13r



SF027 16

2

12

2121

r

qkqF

Solution: q1=1.2x10-6 C, q2=3.7x10-6 C, q3=2.3x10-6 C,

r 12=15x10-2 m, r 13=10x10-2 m

– By applying the Coulomb’s law equation :

Magnitude of F 21:

N 781F 21 .

31F

21F

582q1q

+

3q

-

-

12r 13r

22

6 6 9

2110 x15

10 x7 310 x2110 x09F

)(

).)(.)(.(

k



SF027 17

Magnitude of F 31:

• Construct a table to represents x and y-component for allforces exerted on q1.

Vector sum the x-comp. and y-comp. :

2

13

3131

r

qkqF

N 482F 31 .

Force x-component(N) y-component(N)

21F

21F

0

31F 58F 31 cos

58F 31 sin

N 09358F F F 3121 x1 .cos

N 10258F 0F 31 y1 .sin



SF027 18

8.325

2.34tan

1

1

F F

x

y

from the x-axis anticlockwise

N 783F 1 .

2

y1

2

x11 F F F

The magnitude of resultant electric force exerted on q1 :



SF027 19

1. Two point charges are placed on the x-axis as follows :

Charge q1 = +4.00 nC is located at x = 0.200 m, charge q2 = +5.00 nCis at x = -0.300 m. Find the magnitude and direction of the totalelectric force exerted by these two charges on a negative point charge

q3 = -6.00 nC that is placed at the origin. (Young & freedman,pg.829,no.21.20)(Given 0 =8.85 x 10 -12 C 2 N -1 m -2 )

Ans. : 2.4 N to the right

EXERCISE



SF027 20

2. Four identical point charges (q = +10.0 C) are located on the corners

of a rectangle as shown in figure below.

The dimension of the rectangle are l = 60.0 cm and w = 15.0 cm.

Calculate the magnitude and direction of the resultant electric force

exerted on the charge at the lower left corner by the other threecharges. (Serway & Jewett, pg. 735, no. 57)

(Given 0 =8.85 x 10 -12 C 2 N -1 m -2 )

Ans. : 40.9 N at 263 from positive x-axis.

q

l

+ +

++

w

q

qq

EXERCISE



SF027 21

1.2 ELECTRIC FIELD

LEARNING OUTCOMES:

a) Define Electric Field

b) Define and use Electric Field Strength

c) Sketch the electric field lines of isolated point charge,

two charges and uniformly charged parallel plates

d) Sketch the electric field strength diagram anddetermine electric field strength E for a system of

charges

oq

F E



1.2.1 Electric Field

• Definition – is defined as a region ofspace around isolated chargewhere an electric force is experienced if

a positive test charge placed in theregion.

• Electric field around charges can berepresented by drawing a series of lines.These lines are called electric field lines(lines of force).

• The direction of electric field is tangent to

the electric field line at each point.



SF027 23

• The electric field strength at a point,

Definition – is defined as the electric (electrostatic) force per unitpositive test charge that acts at that point in the same

direction as the force.

Mathematically,

E

E

0q

F E

forceelectrictheof magnitude:F

where

chargetestof magnitude:0q

strengthfieldelectrictheof magnitude: E

1.2.2 Electric Field strength



Electric field strength

• is a vector quantity.

• Its units are N C-1 or V m-1.

Since2

0

r

kqqF

, then the equation above can be written as

0

2

0

qr

kqq

E

2r

kq E or

2

0r 4

q E

chargepointisolatedof magnitude:q chargepointisolatedandpointebetween thdistance:r

where

1 2 3 El i fi ld li



SF027 25

1.2.3 Electric field lines

i. Single positive charge ii. Single negative charge

(the lines point radially outward fromthe charge)

(the lines point radially inwardtoward the charge)

+q

-q

Fielddirection

(a) Isolated point charge



(b) Two charges

i.) Two equal point charges of opposite sign, +q and -q

Field direction

(the lines are curvedand they are directedfrom the positive

charge to thenegative charge.



+q +q X Field direction

(point X is neutral point )

is defined as a point(region) where the totalelectric force is zero.

It lies along the verticaldash line.

ii.) Two equal positive charges, +q and + q

iii ) Two opposite unequal charges +2q and q



SF027 28

+2q-q

iii.) Two opposite unequal charges, +2q and – q

(note that twice as many linesleave +2q as there are linesentering –q, – number of lines is

proportional to magnitude ofcharge.)

Field direction



The rules of drawing electric

field lines:

– The field lines indicate the direction of the electricfield (the field points in the direction tangent tothe field line at any point).

– The lines are drawn so that the magnitude of electric

field is proportional to the number of lines crossingunit area perpendicular to the lines. The closer thelines, the stronger the field.

– Electric field lines start on positive charges andend on negative charges, and the number starting

or ending is proportional to the magnitude of thecharge.

– The field lines never cross because the electric fielddon’t have two value at the same point.



SF027 31

(a) Two equal negative charges, -q and -q .

(b) Two unequal negative charges, -2q and -q.

-q -q

-q-2q

Exercise

1. Sketch electric field lines for the diagrams below:

• Note :



SF027 32

• Note :

– The direction of electric field strength, E depends on sign ofisolated point charge.

– The direction of the electric force, F depends on the sign of

isolated point charge and test charge. For example

• A positive isolated point charge.

a. positive test charge

b. negative test charge

q)( veq0 E F

r

q)( veq0

E

F

r

• A negative isolated point charge



SF027 33

A negative isolated point charge.

a. positive test charge

b. negative test charge

– In the calculation of magnitude E , substitute the magnitude of

the charge only.

q )( veq0 E

F

r

q )( veq0 E F

r



Example 4 :Two point charges, q1=1 C and q2=-4 C, are

placed 2 cm and 3 cm from the point Arespectively as shown in figure below.

Find

a. the magnitude and direction of the electricfield intensity at point A.

b. the total electric force exerted on q0=-4 C if

it is placed at point A.(Given Coulomb’s constant, k = 9.0 x 109 Nm2 C-2)

+ - 2q1q

cm2 cm3

A

Solution: q1=1 C, q2=4 C, q0=4 C, r1=2x10-2 m, r2=3x10-2 m



SF027 35

Solution: q1 1 C, q2 4 C, q0 4 C, r 1 2x10 m, r 2 3x10 m

a. By applying the equation of electric field strength, the magnitude of

E at point A.Due to q1 :

+ - 2q1q

cm2 cm3

A

1 A E 2 A E

22

9

2

1

11 A

10 x2110 x09

r kq E

)())(.(

113

1 A C N 10 x252 E .


Due to q2 : 9 410x09kq ))((



SF027 36

Due to q2 :

therefore the electric field strength at point A due to the charges isgiven by

b. From the definition of the electric field strength,

thus the total electric force exerted on q0 is given by

222

2

22 A

10 x3

410 x09

r

kq E

)(

))(.(

113

2 A C N 10 x4 E Direction : to the right (q2)

2 A1 A A E E E

0

A A

q

F E

1313

A 10 x410 x252 E .

113

A

C N 10 x256 E .


A0 A E qF

N 10 x52F 14

A .

).)(( 13

A 10 x256 4F

Direction : to the left (q1)

EXERCISE



SF027 37

EXERCISE1. Find the magnitude of the electric field at point P due to the four

point charges as shown in the figure below if q=1 nC and d=1 cm.

2. Find the magnitude and direction of the electric field at the centre ofthe square in figure below if q=1.0x10 -8 C and a= 5cm .

(Given 0 =8.85 x 10 -12 C 2 N -1 m -2 )

Ans. : zero.

(Given 0 =8.85 x 10 -12 C 2 N -1 m -2 ) (HRW. pg. 540.13)

Ans. : 1.02x105

N C-1

,

upwards.



SF027 38

1.3 CHARGE IN A UNIFORM

ELECTRIC FIELD

LEARNING OUTCOMES:

a) Explain quantitatively with the aid of a

diagram the motion of charge in a uniform

electric field

1 3 Motion of Charged Particles in a Uniform



SF027 39

• Consider a stationary particle of charge q0 and mass m is placed in a

uniform electric field E , the electric force F e exerted on the charge is

given by

• Since only electric force exerted on the particle, thus this force

contributes the net force, F and causes the particle to accelerate.

• According to Newton’s second law, then the magnitude of the

acceleration of the particle is

• Because the electric field is uniform (constant in magnitude anddirection) then the acceleration of the particle is constant.

1.3. Motion of Charged Particles in a Uniform

Electric Field

E qF 0e

maF F e ma E q0

m

E qa 0



(a) Stationary Charge

Positive stationary charge

E

eF

a

-Consider a stationary particle of charge q0 and mass m is placed in a

uniform electric field E ,

- Electric force F e exerted on the point charge, q0 is given by

E qF 0e

Force experienced by charge is in the same direction as electric field,E.

- q0 can be either positive or negative charge.

For positive charge:

= ma



E

eF

a

For negative charge:

Force experienced by charge is in the opposite direction aselectric field, E.

Negative stationary charge

( )



Consider a charge +q enters the uniform electric field with a velocity

perpendicular to direction of electric field lines.

The positive charge will be deflected and moves along a parabolic path towardsthe negative plate . If the upper plate is positively-charged and the lower is

negatively-charged, then the electric field E will be directed downward.

The positive charge moves under the influence of the electric force which is at the

same direction as electric field lines

(b) Charge moving

perpendicularly to the field

E qF 0

Fe

= may



SF027 43

The negative charge will deflected and moves along a parabolicpath towards the positive plate. The negative charge movesunder the influence of the electric force which is oppositedirection to the electric field lines.

E qF 0e

Consider a charge -q enters the uniform electric field with avelocity perpendicular to direction of electric field lines.

Fe

= may

( ) h ll l



• The electric force on the positive

charge is in the same direction as to

its motion.• The positive charge accelerates along

a straight line.

(c) Charge moving parallel to

the field

v



How about negative charge?

• The electric force on the negative

charge (eg. electron) is in the

opposite direction to its motion.• The negative charge decelerates

along a straight line.

Fe

v

(d) Charge in dynamic equilibrium



SF027 46

• Dynamic equilibrium means the charge moves with constant velocity.

• The positive charged particle will experiences the electric force F E which

is downwards with magnitude qE and the magnetic force F B which

is

upwards with magnitude Bqv (see fig. above).

• If the particle travels in a straight line with constant velocity hence theelectric and magnetic forces are equal in magnitude.

• Only particles with this constant speed can pass through without being

deflected by the fields.

E

v

X X X X X X

X X X X X X

X X X X X X

X X X X X X

B

v

v

BF

E F

(d) Charge in dynamic equilibrium

positive charged particle

for negative charged



for negative chargedparticles.(eg:electron)

Dynamic equilibrium for negative charged particle

E

v

-

X X X X X X

X X X X X X

X X X X X X

X X X X X X

B

v

- v

-

FE

FB

1 4 ELECTRIC POTENTIAL



SF027 48

1.4 ELECTRIC POTENTIAL

OBJECTIVES:

a. Define Electric Potential

b. Define and sketch equipotential lines and surfaces of an isolated charge and a uniform

electric field

c. Use for a point charge and a system of charges.

d. Calculate potential difference between two points

e. Use for uniform E.

f. Deduce the change in potential energy between two points in electric field

g. Calculate potential energy of a system of point charges

r

QV

o 4

d

V E

U

23

32

13

31

12

21

r

qq

r

qq

r

qqk U

V qU

0

BA AB A B

W V V V q

• Electric potential, V of a point in the electric field



SF027 49

Definition – is defined as the work done in bringing positive testcharge from infinity to that point in the electric field.

or

• Since

0q

W

V

then the equation above can be written as

donework :W

chargetest:0q

where

0

0

qr

kqq

V

orr

qV

o 4

r

kqV

r

kqqW 0

chargepoint:qchargepointthepoint withebetween thdistance:r where

spacefreeof typermittivi:0).( 21212

0 m N C 10 x858



SF027 50

• Electric potential is a scalar quantity, so the resultant electric potentialis equal to the algebraic sum of the individual potential.

• The S.I. unit for electric potential is the Volt (V) or J C-1.

• Note :

– The electric potential of a charge at infinity is ZERO . – The electric potential energy of a positively charged particle

increases when it moves to a point of higher potential.

– The electric potential energy of a negatively charged particle

increases when it moves to a point of lower potential.



– Since charge q can be positive or negative, theelectric potential can also be positive or

negative.

– If the value of work done is negative – workdone by the electric force (system).

– If the value of work done is positive – workdone by the external force or on thesystem.

– In the calculation of V , the sign of the chargemust be substituted in the equation of V .

1.4.2 Equipotential Surface



SF027 52

q p• Definition – is defined as a surface where all points on the surface

that have the same electric potential.

• Figures 3.9a and 3.9b are example of the equipotential surface.

– The dashed lines represent the equipotential surface (line).

– The equipotential surfaces (lines) always perpendicular to theelectric field lines passing through them.

Fig. 3.9a : a uniform electric fieldproduced by an infinite sheet ofcharge

E

C

A

B

Fig. 3.9b:a point charge

E

C

A

B

• From the figures,CBA V V V



SF027 53

then the work done to bring a test charge from B to A is given by

EXERCISE:

At a certain distance from a point charge, the magnitude of the electricfield is 500 V m-1 and the electric potential is -3.00 kV. Calculate

a. the distance to the charge.

b. the value of the charge. (Serway & Jewett,pg.788,no.17)

(Given 0 =8.85 x 10 -12 C 2 N -1 m -2 )

Ans. : 6.00 m, -2.00 C

C B A VVV

0W BA No work is done in moving a charge

along an equipotential surface.

0 0 ( ) BA AB A BW q V q V V

1.4.3 Potential Difference



SF027 54

• Potential difference between two points in an electric field,

Definition – is defined as the work done in bringing a positive testcharge from a point to another point in the electricfield.

• From the figure 3.8a, the potential difference between point A and B,

V AB is given by

and

or

A.pointtoBpointfrom

chargetestpositivebringingindonework : BAW where

state)A(finalpointatpotentialelectric: AV state)B(initialpointatpotentialelectric: BV

chargetest:0q

0

BA AB

W V

q

AB A BV V V

0 0 0 0( ) BA AB A B A B A B

W q V q V V q V q V U U

• Note :



SF027 55

– If the positive test charge moving from point A to point B, thus thepotential difference between this points is given by

therefore

B.pointA topointfromchargetestpositivebringingindonework : ABW

where

ApointandBpointbetweendifferencepotential: AB

V

BA AB V V

0

AB

BA B A

W V V V

q

BA

V

Example 5 :



Example 5 :Figure below shows a point A at distance 10 m

from the positive point charge, q=5C.

Calculate the electric potential at point A anddescribe the meaning of the answer.


+q A

m10



SF027 57

Solution: q=5 C, r=10 m

By applying the equation of the electric potential at a point,

)(

))(.(

10

510 x09

r

kqV

9

A

19

A C J V 10 x54V @.

Meaning : 4.5 x 109 joule of work is done in bringing 1 C positivecharge from infinity to the point A.

Example 6 :



Example 6 :Two point charges, q1=+0.3 C and q2=-0.4 C areseparated by a distance of 6 m as shown in

figure below.

Calculatea. the electric field strength andb. the electric potentialat point A ( 3 m from the charge q1).(Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2)

+ - 2q1q A

m6



SF027 59

Solution: q1=+0.3 C, q2=-0.4 C

a. By applying the equation of electric field strength, the magnitude of

E at point A.

Due to q1 :

+ - 2q1qA

m3r 1 m3r 2 1 A E

2 A E

2

9

2

1

11 A

3

3010 x09

r

kq

E )(

).)(.(

18

1 A C N 10 x3 E


Due to q2 : 9

2 4010 x09kqE

).)(.(



SF027 60

therefore the electric field strength at point A due to the charges isgiven by

b. By applying the equation of electric potential, the value of V at point

A is

22

2

2 A3r

E )(

18

2 A C N 10 x4 E Direction : to the right (q2)

2 A1 A A E E E

2 A1 A A V V V

88

A 10 x410 x3 E

18

A C N 10 x7 E


2

2

1

1

2

2

1

1 A

r

q

r

qk

r

kq

r

kqV

3

40

3

3010 x09V

9

A

...

V 10 x3V 8

A

Example 7 :



Example 7 :

Two point charges, q1=+12 nC and q2=-12 nC

are separated by a distance of 8 cm as shown infigure below.

Determine the electric potential at point P( 6 cmfrom the charge q2).


1q + - 2q

P

m8 c

m6 c

Solution: q1=+12x10-9 C, q2=-12x10-9 C



SF027 62

1

q+ -

2

q

P

m10 x82

m10 x6 r 2

2 m10 x10r 2

1

By applying the equation of electric potential, the value of

V at point P is

2P1PP V V V

2

2

1

1

2

2

1

1P

r

q

r

qk

r

kq

r

kqV

V 720V P

Exercise:



SF027 63

Exercise:

1. Four point charges are located at the corners of a square that is 8.0 cmon a side. The charges, going in rotation around the square, are q, 2q, -3q and 2q, where q = 4.8 C as shown in figure below.

q q2

q2 q3

cm8

Find the electric potential at the centre of the square.(Given 0 =8.85 x 10 -12 C 2 N -1 m -2 )

Ans. : 1.53 x 106 V.

Example 8 :



Example 8 :Two point charges q1=+2.40 nC and q2=-6.50 nC are

0.100 m apart. Point A is midway between them,point B is 0.080 m from q1 and 0.060 m from q2 asshown in figure below.

Find

a. the electric potential at point A,

b. the electric potential at point B,

c. the work done by the electric field on a charge of 2.5 nC thattravels from point B to point A.

(Young & freedman,pg.900,no.23.21)

1q+ -

2q

B

A

m060.0m080.0

m050.0 m050.0

B



SF027 65


Solution: q1=+2.40x10-9 C, q2=-6.50x10-9 C,

r 1A=r 2A=0.050 m, r 1B=0.080 m , r 2B=0.060 m

a. By applying the equation of electric potential, the value of V at point

A is

A2 A1 A V V V

V 738V A A2

2

A1

1 A

r

kq

r

kqV

1q+ -

2q A

m060.0m080.0

m050.0 m050.0

b. By applying the equation of electric potential, the value of V at point



SF027 66

B is

c. Given q0=2.50x10-9 C

The work done in bringing charge, q0 from point B to point A is

given by

B2 B1 B V V V

V 705V B B2

2

B1

1 B

r

kq

r

kqV

J 10 x258W 8

BA

.

)( B A0 BA V V qW 0

BA AB

W q V

Example 9 :



Example 9 :

A test charge q0=+2.3x10-4

C is 5 cm from a pointcharge q. A work done of +4 J is required toovercome the electrostatic force to bring the testcharge q0 to a distance 8 cm from charge q.

Calculate :

a. the potential difference between point 8 cmand 5 cm from the point charge, q.

b. the value of charge q.c. the magnitude of the electric field strength forcharge q0 at point 5 cm from the charge q.


Solution: q0=+2.30x10-4 C



SF027 68

a. Given W AB= +4J ,

From the figure above, r A= 5x10-2 m, r B= 8x10-2 m By applying the equation of potential difference, the value of V BA is

b. The electric potential at point A due to point charge, q :

)(

)(2

9

A

A10 x5

q10 x9

r

kqV

q10 x81V 11

A .

q B A

m10 x5 2

m10 x8 2

eF

F

0

AB BA

W

V q41.74 10

BAV V

The electric potential at point B due to point charge, q :

109k9 )(



SF027 69

The potential difference between point A and B is

c. By using the equation of electric field strength, thus

C 10 x582q 7 .

qq x 11114 108.110125.11074.1

2 A A r

kq E

and

15

A C N 10 x299 E .

q10 x125110 x8

q10 x9

r

kqV

11

2

9

B

B .)(

)(

41.74 10 BAV V BA B AV V V

1.4.5 Relation Between V and EC id iti t t h l d iti i t



SF027 70

• Consider a positive test charge, q0 placed near a positive point

charge, q. To move q0 towards q by a small displacement (r ), work

done (W ) must be expended as shown in figure below .

• The work done by the external force F is given by

and

0r F W cos

V qW 0

eF

F

Δr

q +0q

r q

F V

0

e

eF F r F W e

Since then

r V E

and E q

F

0

e

r E V or

differencepotential:V where

e)nt(distancdisplacemeinchange:r

strengthfieldelectric: E

• In the limit when r approaches zero,

V



SF027 71

– The negative sign indicates that the value of electric potentialdecreases in the direction of electric field.

– is known as the electric potential gradient. It can be obtained

from the gradient of a V against r graph.

• An alternative unit for electric field strength, E is volts per meter where

• . The electric field produced by a pair of flat metal plates, one of which

is earthed and the other is at a potential of V is uniform. This can be

shown by equally spaced lines of force in figure 1.

dr

dV

E

11 mV 1C N 1

dr

dV

r

V E

0rlimit

d V



SF027 72

– The V against r graph for pair of flat metal plates can be shown in

figure 2.

– From the figure 2,

• The graph is a straight line with negative constant gradient,thus

V

Fig.1

0V

r

V

0 d Fig. 2

)(

)(

0d

V 0

Δr

ΔV E

d

V

E or

Ed V Uniform E

1 4 6 Changes in potential



1.4.6 Changes in potential

energy , ∆ U

• Electric potential energy, U of a point

charge, q which is at a distance of r

from point charge,q is

U = qV

• If there is changes in potential energy

between 2 points in electric field, ∆ U = q ∆ V

1 4 7 Electric Potential Energy of System of



SF027 74

1.4.7 Electric Potential Energy of System of

point charges

• The electric potential energy of the system of point charges is the work

done to bring all the charges from infinity to the points where the

charges are placed.

• In the system of charges, suppose there were originally no charges at the

points A, B and C as in the figure above.

C

q3

B

q2

A

q1

r12

r23

r13

To set up the system,



SF027 75

- The charge q1 is brought from infinity and placed at point A. Since there

where originally no charges, the charge q1 doesn’t experience any electric

force when it is brought from infinity, Fe = 0

Work done,

- With the charge q1 fixed at the point A, an electric field is produced. Hence

the work done to bring the charge q2 from infinity to the point B which is at

the distance of r 12 from charge q1 is,

Work done,

- With the charge q1 fixed at point A and q2 fixed at point B, the electric

potential at point C ,

0 A A U W

12

21

r

qkqU W B B

23

2

13

1

r

kq

r

kqV C

- Hence the work done to bring charge q3 from infinity to the point C is



g g q3 y p

Work done,

- Therefore, the electric potential energy of the system,

23

2

13

133

r

kq

r

kqqV qU W C C C

321 U U U U

23

2

13

1

3

12

210

r

q

r

qkq

r

qkq

23

32

13

31

12

21

r

qk q

r

qk q

r

qk q

323121

r

qq

r

qq

r

qqk U

unit editted

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