Chapter 3

(a) q1q2 < 0 (b) q1q2 > 0 Figure 3.1

PHY193 - Basic Physics For Engineers II

CHAPTER 3 - ELECTRIC POTENTIAL

3.0 Electric potential energy

- Electric potential energy is the energy stored in an electric field. The change in potential energy when charged particles move around in an electric field is equal in magnitude but opposite in sign to the work done by the electric field.

ΔU E=−W field (3. 1 )

From equation 3.1,

i) Wfield = +ve, UE = -ve. The electric potential energy decreases because it is used to move the

particles.

ii) Wfield = -ve, UE = +ve. The electric potential energy increases.

- The equation for electric potential energy between two point charges is given by

U E=kq1q2

r (3 .2 )

where r is the separation distance between q1 and q2. Figure 3.1a shows the change in

UE when q1 and q2 have opposite signs. Figure 3.2b shows the change in UE when q1 and q2 have the same signs.

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PHY193 - Basic Physics For Engineers II

3.1 Relation between electric potential and electric potential energy

- Electric potential, V is defined as the electric potential energy per unit charge.

V=

U E

q (3 .3 )

Electric potential is a scalar quantity. The SI unit for electric potential is volt(V).

- When a point charge q moves from point A to point B in an electric field, it moves through a potential difference

ΔV=V B−V A

=ΔU E

q(3 .4 )

- If two charges q1 and q2 are separated at a distance r from each other, the electric potential at a distance r from q1 is given by

V=kq1q2

rq2

¿kq1

r (3 .5 )

The electric potential at a point P due N point charges is the sum of the potentials due to each charge.

V P=∑

i=1

N

V i=∑i=1

N

( kq i

ri ) (3. 6 )

where ri is the distance from the ith point charge, qi to point P.

Example 3.1

Charge q1 = +4.0C is located at (0,3) cm, charge q2 = +2.0C is located at (1,0) cm and charge q3 = -3.0C is located at (2,2) cm.

a) Find the electric potential at point A(0,1) cm due to the three charges.

b) A point charge q = -5.0 nC moves from a great distance to point (0,1) cm. What is the change in the electric potential energy?

33

y

xq2

q3

q1

A

B C

qA = 3C

qB = 1C qC = -2CFigure 3.2

PHY193 - Basic Physics For Engineers II

Solution

a)

The distances of each charge from point (0,1) are

r1 = 2.0 cm, r2=√12+12=1. 414 cm

, r3=√12+22=2 .236 cm

The potential at point (0,1) is the sum of the potentials due to each point charges,

V total=∑kqiri

=( 8. 99×109 )[4 μ0 .02

+2μ0 .01414

+−3 μ0 .02236 ]

=+ 1. 863×106 V

b) The change in potential energy, UE = qV

V = Vfinal - Vinitial = VA - 0 (The charge is assumed to start from r = , V = 0) = VA

UE = (-5.0 x 10-9)(1.863 x 106) = -9.3 x 10-3 J

Example 3.2

Figure 3.2 shows three point charges placed at the corners of an equilateral triangle ABC whose sides are 20 cm long. Find the electric potential of these source charges at the midpoint of line AD.

D

34

Figure 3.3

PHY193 - Basic Physics For Engineers II

Solution

Take the midpoint of AD as point E.

AB=√ AD2+DB2

0 .2=√ AD2+0 .12

AD=√0 .22−0 . 12=0 .1732 m

⇒ AE=DE=12AD=0 .0866 m

BE=√BD 2+DE2=√0 .12+0 .08662

¿0 . 1322 mBE=CE

The electric potential at point E is

V E=V A+V B+V C

¿ (8 .99×109) [3 μ0. 0866+1μ

0 .1322+−2μ

0. 1322 ]¿2 .43×105 V

3.2 Equipotential surface

- Equipotential surface is defined as a surface on which the potential is constant. The

electric field lines(direction of E ) is always perpendicular through these surfaces. For conductors in electrostatic equilibrium, the surfaces are equipotential surfaces.

- Figure 3.3 shows an example of a group of equipotential surfaces. If a charged particle moves between two points on an equipotential surface, no work is done on the particle. This is illustrated by paths I and II.

If two charged particles move in different paths but between the same pair of equipotential surface, work done on both particles has the same value. This is illustrated by paths III and IV. The work done on the particle is independent on the path taken by the particles.

35

Figure 3.5

PHY193 - Basic Physics For Engineers II

- Figure 3.4 shows the equipotential surfaces of several charge distributions.

(a) Single positive charge

(b) Two charges of same magnitude but different sign

(c) Uniformly charged planar surface

Example 3.3

When an electron moves from A to B along an electric field line in Figure 3.5, the electric field does 3.94 x 10-19 J of work on it. What are the electric potential differences

a) VB - VA b) VC - VA c) VC - VB

Solution

From equation (3.1), UE = - W = - 3.94 × 10−19 J.

a ) V B−V A=ΔU E

q=−3 . 94×10−19

−1 . 60×10−19=2.46 V

b) VC − VA = VB − VA = 2.46V Movement occur between same pair of equipotential line.

Figure 3.4

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PHY193 - Basic Physics For Engineers II

c) VC − VB = 0 Since C and B are on the same equipotential line.

3.3 Potential in a uniform electric field

- In a uniform electric field, the field lines are equally spaced parallel lines. Since equipotential surfaces are perpendicular to field lines, the equipotential surfaces are a set of parallel planes (See Figure 3.4c). The potential decreases from one plane to the next in the direction of electric field.

- Work done by the electric field to move a charged particle a distance d in the field is

W field=FE d=qEd (3.7)

The change in electric potential energy is

ΔU E=−W field=−qEd

The change in potential is

ΔV=ΔUE

q=−Ed

(3.8)

The -ve sign tells us that the electric potential decreases in the direction of the electric field.

Example 3.4

An electron is released in a constant electric field of magnitude 150 N C-1 and of direction along the -ve y axis. What is the change in the electric potential energy of the electron when an electrostatic force causes it to move vertically upward through a distance d = 520 m?

Solution

The change in the electric potential energy, UE is related to the work done by the electric field on the electron by equation (2.1).

In a constant electric field, the electron is subjected to a constant electric force, F = qE.

Work done by a constant force is given by W=F⋅d=Fd cosθ

Work done by the electric force is W=qEd cos θ

The field is directed downward, that is opposite to the displacement of the electron = 180o.

Work done by the electric force on the electron,

W=(−1 .6×10−19 ) (150 ) (520 ) cos (180o)=1. 2×10−14 J

From equation (3.1),

UE = -W = -1.2x10-14 J.

37

Area A

d

+ q

- q

Figure 1.20

PHY193 - Basic Physics For Engineers II

The –ve sign indicates that the energy of the field decreases. This is because 1.2 x 10-14 J of electric potential energy is used to move the electron in the field.

3.4 Capacitor

- A capacitor is a device that is used to store electric charge. A basic structure of a capacitor consists of a pair of conducting plates separated by an insulator called dielectric. The “ “ symbol is used to denote a capacitor in an electric circuit.

- The measure of extent to how much electric charges can be stored in a capacitor is the capacitance of a capacitor. It is defined as

C=QV (3.9)

whereC = The capacitance ( unit = Farad, symbol = F)Q = The magnitude of charge on either plate ( unit = Coulomb)V = The potential difference between the plate ( unit = Volt, symbol = V )

Example 3.5

(a) If the charge on a capacitor is 50C when the voltage across it is 25V, what is the capacitor’s capacitance?(b) If the voltage across this capacitor is increased to 100V, what is the charge on the capacitor?

Solution

(a ) C=QV

=50×10−6C25V

=2×10−6C

(b ) Q=CV=(2×10−6 F )(100V )=200×10−6C

3.4.1 Parallel plate capacitor

- Figure 3.6 shows the arrangement of a parallel plate capacitor where the magnitude of the charge on the surface of either plate is Q and the potential difference between them is V. Let the area of the plate be A and the plate separation be d. The space between the plates are filled by insulators known as the dielectric material.

- If we assume that the dielectric material between the plates is free space / air / vacuum, having the permittivity constant, 0, from the Gauss Theorem, the electric field strength between the plates is given by

E= σε0

= Qε 0 A (3.10)

where

σ=QA = Charge per unit area of the plate/Surface charge density

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PHY193 - Basic Physics For Engineers II

- From equation (3.9), the capacitance of a parallel plate capacitor is given by

C=QV

=ε0 EA

V (3.11)But

E=Vd (3.12)

By substituting Eq. 3.12 in Eq. 3.11, the capacitance of a parallel plate capacitor is given by

C=ε0 A

d (3.13)

- Equation 3.13 indicates that the capacitance of a parallel plate capacitor depends only on its plate area and plate separation.

Example 3.6

(a) What is the capacitance of a parallel plate capacitor that has square plates with lateral dimensions of 122 mm on one side, a plate separation of 0.24 mm and vacuum between the plates?

(b) What is the charge on the capacitor if the potential difference across it is 45V?

Solution

(a ) C=ε0 A

d=

(8 .85×10−12F /m)(0 .122m)2

2 .4×10−4 m=5 .5×10−10 F

=550 pF

(b ) Q=CV=(550×10−12F )( 45V )=25×10−9 C=25nC

Example 3.7

39

PHY193 - Basic Physics For Engineers II

The plates of a parallel plate capacitor are in vacuum, 5.00 mm apart and 2.0 m2 in area. A potential difference of 10kV is applied across the capacitor. Compute(a) The capacitance of the capacitor.(b) The charge on each plate.(c) The magnitude of the electric field in the space between the plates.

Solution

(a ) C=ε0 A

d=

(8 . 85×10−12F /m) (2 .00m2)5 .00×10−3 m

=3 .54×10-9=3 .54nF

(b ) Q=CV=(3 . 54×10-9F ) (10×103V )=35 .4 μC The plate at higher potential has a charge of +35 .4 μC and the other plate has a charge of - 35 .4 μC .

(c ) E=σε0

=Qε0 A

=35 . 4×10−6C

(8 .85×10−12F ) (2 . 00m )=2 .00×106 N /C

or

E=Vd

=10×103V5 .00×10−3m

=2 . 00×106 V /m

3.4.2 Effect of A Dielectric Material

- In capacitors, dielectric materials are used

(1) To increase the capacitance of a capacitor.

(2) To allow the capacitor to be built in practical shapes and sizes. For example, paper is used as the dielectric material in a capacitor made of flexible plates of aluminum foil rolled into a cylinder.

(3) To limit the potential difference that can be applied between the plates to a certain value Vmax , called the breakdown potential. If this value is exceeded, the dielectric material will breakdown and form a conducting path between the plates. Every dielectric material has a characteristic dielectric strength, which is the maximum electric field that can be tolerated without breaking down.

- The dielectric properties of dielectric materials are characterized by the dielectric constant, K of the material. Table 3 shows the value of K for several dielectric materials.

Table 3 – Values of K and dielectric strength for several dielectric material

Material Dielectric constant, K Dielectric strength(kV/mm)Vacuum 1Air (STP) 1.000576 3Polystyrene 2.6 24Paper 3.5 16Pyrex 4.7 14

- The dielectric constant of a dielectric material is defined by

40

PHY193 - Basic Physics For Engineers II

K= CC0 (3.14)

where C = The capacitance when the dielectric material is between the plates

Co = The capacitance when there is air or vacuum between the plates.

- If no extra electric charge is supplied to the capacitor when the dielectric material is added,

K=CCo

=

Qo

VQo

V o

=V o

V (3 .15)

The presence of a dielectric material reduces the potential difference between the plates by a factor of K where

V = The potential difference between the plates with the dielectric Vo = The potential difference between the plates without the dielectric

- Because the potential difference between the plates is reduced by a factor K when a dielectric is present, the electric field between the plates is reduced by the same factor. If Eo is the vacuum value and E the value with the dielectric, then

Eo

E=K

(3.16)

- From equation 2.14,

K= CC0

=Kεo A /dεo A /d

= εε o

(3 .17)

where

= The permittivity of the dielectric material = Kεo

o = The permittivity of free space/ air / vacuum

- In terms of , we can write

E=σε (3.18)

and

C=KC o=KεoAd

=εAd (3.19)

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PHY193 - Basic Physics For Engineers II

Example 3.8

A parallel plate capacitor is initially charged to a potential difference of 3kV. After a sheet of dielectric material is inserted between the plates, the potential difference decreased to 1kV. If the plates have an area of 0.2m 2 and are 1.00cm apart, compute(a) The capacitance before the dielectric is inserted.(b) The magnitude of charge on each plate before the dielectric is inserted.(c) The capacitance after the dielectric is inserted.(d) The dielectric constant, K of the material.(e) The permittivity of the material.(f) The electric field before the dielectric is inserted.(g) The electric field after the dielectric is inserted.

Solution

(a ) C0=ε 0 A

d=

(8 .85×10−12F /m ) (2 .0×10−1m2 )1 .00×10−2m

=17 .7×10-11F

(b ) Qo=CoV o=(17 .7×10−11) (3000 )=0 .531×10−6C )

(c ) C=Q0

V=0 .531×10−6C

1 .00×103V=53.1×10−11F

(d ) K=CC0

=53 .1×10−11 F17 .7×10 -11F

=3 .00

(e ) ε=Kε0=( 8.85×10−12F /m ) (3 )=26 .6×10−12F /m

( f ) E=V 0

d=3000V

1 .00×10−2m=3 .00×105 V /m

( g) E=Vd

=1000V1 .00×10−2m

=1 .00×105V /m

or

E=Q0

εA=0 .531×10−6C

(26 .6×10−12F /m ) ( 2.0×10−1m2)=1.00×105V /m

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(a) (b) Figure 3.7

C1 C2 C3

C1

C2

C3

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3.5 Series and parallel capacitor

- Capacitors can be connected in two basic ways, in series and in parallel.

- In series, the capacitors are connected “head to tail” (See Figure 3.7a). In parallel, the capacitors are connected “head to head” and “tail to tail” and all the leads on one side of the capacitors have a common connection (See Figure 3.7b).

- In series, the charges are the same on all plates Q1 = Q2 = Q3 = Q

The sum of voltage drop across all capacitors equals the voltage of the source,

VS = V1 + V2 + V3 Q/Ceq = Q/C1 + Q/C2 + Q/C3

1/Ceq = 1/C1 + 1/C2 + 1/C3 (3.20)

The value of Ceq is the equivalent series capacitance that is the three capacitors in series could be replaced with one capacitor whose capacitance is Ceq.

For any n number of capacitors in series,

1CS

=∑i=1

n1C i (3.21)

- In parallel, the voltage across the capacitors is the same.

V1 = V2 = V3 = VThe total charges is the sum of the charges of individual capacitors,

Qeq = Q1 + Q2 + Q3 CeqV = C1V + C2V + C3V

Ceq = C1 + C2 + C3 (3.22)

The value of Ceq is the equivalent parallel capacitance. That is the three capacitors in parallel could be replaced with one capacitor whose capacitance is Ceq.

For any n number of capacitors in parallel,

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3F

9F

6F

12F

11F

C1

C5

C4

C3

C2

Figure 1.22Figure 3.8

120V

2F

4F3F

X

Y

Z

Figure 1.23Figure 3.9

PHY193 - Basic Physics For Engineers II

CP=∑i=1

n

Ci (3.23)

Example 3.9

Find the equivalent capacitance of the combination shown in Figure 3.8.

Solution

C3 and C4 are in series to each other,

⇒1C34

=1C3

+1C4

=112 μ

+16 μ

=312μ

⇒C34=4 μF

C34 , C1 and C2 are parallel to each other,⇒C1234=C34+ C1+C2=18 μF

⇒1Ceq

=1C1234

+1C5

=118 μ

+19μ

=318 μ

⇒Ceq=6 μF

Example 3.10

From Figure 3.9, calculate a) the equivalent capacitance of the circuit. b) the potential difference across capacitor X, Y and Zc) the charges on each of the capacitor.

Solution

a ) CY and CZ are in parallel to each other⇒CYZ= CY+CZ=6μF

⇒1CT

=1CX

+1CYZ

=13μ

+16 μ

=36μ

⇒CT=2μF

b) For the circuit as a whole, the total charge is QT=CT V T=(2×10−6) (120 )=240μC .

44

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Since CX and CYZ are in series, QX=QYZ=240μC.

If the potential across X is V X , ⇒V X=QX

CX

=240μ3 μ

=80V

ε=V X+V YZ ⇒V YZ=ε−V X=120V−80V=40V

Since CY is parallel to CZ , VY=V Z=40V

c) QX = 240C

QY=CY V Y= (2×10−6 ) (40 )=80 μC

QZ=CZ V Z=( 4×10−6 ) (40 )=160μC

3.6 Energy stored in a charged capacitor

- During the charging of a capacitor, the addition of electrons to the negative plate involves doing work against the repulsive forces of the electrons that are already there. As charges accumulate on the capacitor plates, the battery will have to do increasingly larger amounts of work to transfer additional electrons. The work to overcome the repulsive force is stored in the form of electrical potential energy in the electric field between the plates..

- The electrical potential energy stored in the capacitor is calculated using

U=Q2

2C=1

2CV 2=1

2QV

(3.24)

Example 3.11

A certain parallel plate capacitor consists of two plates, each with area 200 cm2, separated by a 0.4 cm air gap.(a) Compute the capacitance of the capacitor.(b) If the capacitor is connected to a 500V source, compute the energy stored in the capacitor.(c) If a liquid with K = 2.60 is poured between the plates so as to fill the air gap, determine the energy now

stored in the capacitor.

Solution

(a ) C=ε0 A

d=

(8 .85×10−12) (0 .02 )0 .004

=44 pF

(b ) Q=CV=( 44×10−12 ) (500 )=22nC

U=12QV=1

2(22×10−9) (500 )=5.5 μJ

45

X

Y

Figure 1.24Figure 3.10

PHY193 - Basic Physics For Engineers II

(c ) The capacitor will now have a capacitance 2 .60 times larger than before . Extra charges will now flow into the capacitor .T he amount of charg es now stored in the capacitor is

Q'=KQ0=(2.60 ) (22×10−9 )=57nC

The energy stored in the capacitor is U=12

(57×10−9 ) (500 )=14 .25 μJ

Example 3.12

A 5F capacitor(X) is fully charged by a 40V supply. The supply is then disconnected the capacitor. It is then connected across an uncharged 20F capacitor(Y) (See Figure 3.10). Calculate(a) the final potential difference across each(b) the final charge on each(c) the initial and final energies stored by the capacitors.

Solution

(a) The initial charge, Q0 = CXV0 = (40)(5 x 10-6) = 200 C

When X and Y are connected, they are parallel to each other Total capacitance, CT = CX + CY = 5F + 20F = 25 F

The total charge is unchanged, Q0 = QX + QY

Q0 = CTVT 200 x 10-6 = (25 x 10-6) VT

VT = 200/25 = 8 V

(b) When X and Y are connected, the total charge is unchanged.

For Y, QY = CYVT = 8 x (20 x 10-6) = 160 x 10-6 C

Since the total charge is conserved, Q0 = QX + QY

QX = Q0 - QY = (200 – 160) x 10-6 = 40 x 10-6 C

(c) The initial energy, Ui = (1/2)CXV02 = (1/2)(5 x 10-6)(40)2 = 4mJ

The final energy, Uf = (1/2)CXVT2 + (1/2)CYVT

2 = (1/2)(5 x 10-6 + 20 x 10-6)(8)2 = 8 x 10-4 J

46

q1q2

A

B

0.05 m 0.05 m

0.08 m

0.06 m

Figure 1

PHY193 - Basic Physics For Engineers II

Supplementary Problems

A) Electric potential

1. A charge of 28.0 nC is placed in a uniform electric field that is directed vertically upward and that has a magnitude of 4.0 x 104 Vm-1. What work is done by the electric force when the charge moves

a) 0.45 m to the right? b) 0.67 m upward? c) 2.60 m at an angle of 45o below the horizontal?

(Answer: a) 0 J b) 7.5 x 10-4 J c) -2.06 x 10-3 J)

2. a) What is the electric potential at a distance 0.25 m away from a point charge of 1.0 x 10-9 C?

b) How much work is required to bring an identical charge from very far away up to this distance of 0.25 m from the first charge?

(Answer : a) 36 V b) 3.6 x 10-8 J)

3. In Figure 1, two point charges q1 = +2.40 nC and q2 = -6.50 nC are 0.10 m apart. Point A is between them; point B is 0.080 m from q1 and 0.060 m from q2. Finda) the potential at point A.b) the potential at point B.c) the work done by the electric field on a charge of 2.50 nC that travels from point B to

point A.

(Answer: a) -738 V b) -705 V c) 8.25 x 10-8 J )

4. Four point charges q = +3.6C are arranged in a square of edge length 2.0 cm.a) What is the electric potential energy at the center of the square?b) What is the electric potential at the midpoint of one edge?(Answer: a) 9.2 x 106 V b) 9.4 x 106 V)

B) Capacitor

5. A parallel plate capacitor consist of two square plates each of side 25 cm , 3.0 mm apart. If a p.d. of 200 V is applied, calculate the charge on the plates with (i) air (ii) paper of dielectric constant 2.5, filling the space between them.

(Answer : (i) 37 nC (ii) 93 nC)

6. A parallel plate capacitor has a capacitance of 1.5 F with air between the plates. The capacitor is connected to a 12 V battery and fully charged. When a dielectric is placed

47

paper air

Figure 2

3F 6F

F

12 V6F

4F

5F

1000 V

(a) (b) Figure 3

10F

30F

30F

12F

12F

12 V

Figure 4

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between the plates, a potential difference of 5.0 V is measured across the plates. What is the dielectric constant of the material?

(Answer : 2.4)

7. A capacitor is made of two parallel plates , each with an area of 146 cm2. The plates are separated 0.58 mm from each other. Half of the area is filled with paper and half is filled with air (See Figure 2). Calculate the capacitance of the capacitor.

(Answer: 0.50 nF) 8. Calculate the equivalent capacitance in Figure 3a and 3b.

(Answer : (a) 6F, (b) 3.33F )

9. In Figure 4, calculate a) the equivalent capacitance in the circuit. b) the voltage across the 10F capacitor in the circuit.

(Answer: a) 4.84F b) 2.32 V)10. Compute the energy stored in a 60 pF capacitor a) when charged to a potential difference of 2 kV. b) when the charge on the plate is 30 nC.

48

3F 5F

4F

12V

Figure 4Figure 5

PHY193 - Basic Physics For Engineers II

(Answer : (a) 120 J (b) 7.5 J)

11. A parallel plate capacitor having area 40 cm2 and spacing of 1 mm is charged to a potential difference of 600V. Determine(a) the capacitance.(b) the magnitude of charge on each plate.(c) the energy stored in the capacitor. (d) the electric field between the plate.(Answer: (a) 35pF (b) 21nC (c) 6.3J (d) 6.0x105V/m)

12. A parallel plate capacitor is made up of two parallel plates of area 600cm2 separated at 5mm by a layer of wax paper of dielectric constant of 2.0. Compute(a) the capacitance of the capacitor.(b) the energy stored in the capacitor when it is connected across a p.d. of 6V.(Answer : (a) 212.4 pF (b) 3.82 nJ)

13. From Figure 5, calculate(a) the total electrical energy stored in the 5F and 4F capacitors.(b) the electric charge stored in the 3F capacitor.(Answer : (a) 160J (b) 36C)

49