PHY193 Bab 3.Docx

46 ELECTRIC POTENTIAL

PHY 193 Physics For Engineering II UiTM Pulau Pinang

3

E L E C T R I C P O T E N T I A L

3.1 ELECTRIC POTENTIAL Electric Strength can also be defined in units of work and energy. This alternative way of

measuring the strength of an electric field entails finding the potential difference, which exits

between any two points. To visualizing this difference picture an electric field, then trying to

force a test charge in between two points, if the test charge is repelled, we have to do work to

push it into place.

The potential at any point is defined as the work done per unit charge in moving a positive

charge from zero potential to the point. (Let us choose infinity as the reference point for zero

potential)

V = 0q

W A∞ =

0q

Fr=

2

0

0

rq

rkQq

V = r

kQ

V =r

kQ=

0π4

1

∈

r

Q (the electric potential at a distance r from a single point charge Q)

The electric potential at a point caused by a collection at the Point like charges Qi is found

from the principle of superposition:

V = V1 + V2 + V3 + …

V =

1

1

r

kQ+

2

2

r

kQ +

3

3

r

kQ+ …

V = k ( 1

1

r

Q +

2

1

r

Q+

3

1

r

Q+ … )

Electric potential is scalar quantity and measured in Volts (1V = 1J/C)



Potential difference (difference in potential), between two points a and b is measurable. Since the difference in potential energy Va – Vb , is equal to the negative of work Wba , done by the

electric force to move the charge from point b to point a

( )ba a bW q V V= − −

The work done to move the charge is given by

W q V= − ∆

Example 3.1

Two point charges Q1 = +0.015 µ C and Q2 = -0.015 µ C are placed as in a Figure 3.1.

Compute the potential at point A .

Solution

VA = VA1 + VA2

VA =1

1

r

kQ+

2

2

r

kQ

VA = 04.0

)10015.0(109 69 −×× +

14.0

)10015.0(109 69 −×−×

VA = 3375 + (-964.3) = +2410.7V

Exercise 3.1

As in an example 3.1 and Figure 3.1. Compute the potential at point B and C.

Q2 Q1

B A

C

10 cm 10 cm

4 cm 6 cm 4 cm

Figure 3.1



Example 3.2

Two point charges are arranged as in a Figure 3.2. Find the

(a) potential at point X (b) potential at point Y

(c) difference potential between point X and Y

(d) work done to move a point charge q = 5µC from X to Y

Solution

(a) VX = VX1 + VX2 = 450000 V + (-630000 V) = -180000 V

(b) VY = VY1 + VY2 = 360000 V + (-420000 V) = -60000 V

(c) VXY = VY – VX = -60000 - (-180000) = 120000 V

(d) q

WXY= - VXY

WXY = - VXY (q ) = - (120000 ) (5 X 10-6)

= - 0.6 J

Exercise 3.2

From the Figure 3.2 above, find the work done to move a point charge q = -8µC from X

to infinity

Y

X

10 cm

20 cm

20 cm

Q2 = -14µC

Q1 = +20µC

Figure 3.2



3.2 EQUIPOTENTIAL LINES/SURFACE

The electric potential can be representing diagrammatically by drawing equipotential lines or,

in three dimensions, equipotential surfaces. An equipotential surface is one on which all point

are at the same potential.

That is, the potential difference between any two points on the surface is zero, and no work is

required to move a charge from one point to other. An equipotential surface must be

perpendicular to the electric field at any point.



Example 3.3

When an electron moves from A to B along an electric field line in Figure 3.3, 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

a)

-19

-19

-

- ( - )

--

3.94 10

-(-1.60 10 )

2.46

AB

B A

ABB A

W q V

q V V

WV V

q

x

x

V

= ∆

=

=

=

=

b) VC − VA = VB − VA = 2.46V

⇒ Movement occur between same pair of equipotential line.

c) VC − VB = 0

⇒Since C and B are on the same equipotential line.

Figure 3.3



Example 3.4

An electron is released in a constant electric field of magnitude 150 NC-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.

In a constant electric field, the electron is subjected to a constant electric force,

F = qE. (i)

Work done by a constant force is given by

)ii(cosFddFW θ=•=vv

Combining equation (i) and (ii),

)iii(cosqEdW θ=

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

⇒ θ = 180o.

The work done on the electron,

( )( )( ) ( )J102.1

180cos520150106.1W

14

o19

−

−

×=

×−=

From equation,

∆UE = -W = -1.2x10-14

J.

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.



Tutorial 3.1

1. Two point charges Q1 and Q2 lay 6m apart on a straight line as shown in Figure

3.5. Find the potential at the point midway between two charges.

(Ans : 21 kV)

2. Two point charges Q1 = +20 µC and Q2 = -10µC are 20 cm as shown in Figure

3.6. Find

(a) the potential at point x

(b) the potential at point y

(c) potential difference between points x and y.

(d) work done to move a charge 5µC from x to y.

(Ans : 0 V, 60 kV, - 0.3 J)

3. Two point charges Q1 and Q2 are arranged as shown in Figure 3.7. Find

(a) the potential at point A and B

(b) work necessary to move a +20 µC test charge from point A to B

(c) work done to move a 5µC test charge from point B to infinity

(Ans : 0 V, -1.44X 105 V, +2.88 J, -0.72 J )

Q3 = +2µC

Q3 = +5µC

6 m

Figure 3.5

Q2 Q1

20 cm 20 cm 10 cm

X Y

Figure 3.6

Figure 3.7

Q2 = -2µC

A B 30 cm 20 cm 10 cm

Q1 = +2µC



4. Question 4 is referring to the Figure 3.8,

(a) Calculate the electric force on Q1 due to other charges.

(b) Find the electric field at origin (0,0)

(c) Calculate the electric potential at origin (0,0)

(Ans : (a) 0.072N, 16.93(3rd quarter) ; (b) 27.75 x 103NC-1, θθθθ = 54.20 1st

quarter ; (c) -37.53 x 103V)

5. Question 5 is referring to the Figure 3.9

(a) draw the electric force vector on Q3 due to charge Q2 and Q1

(b) find the resultant force on Q3 .

(c) find the electric potential at point A and P

(d) find work done to move a charge 2.0µC from point P to A.

(Ans :(b) 72.94kN, 17.94οοοο (4

th quarter);(c) 3.825 x 10

7 V (c) 3.6 x 10

7 V; (d) -4.5 J)

Y (m)

Q2 = +5µC

Q3 = -5µC

Q1 = -10µC

(-2,0) (3,0)

(0,2)

X (m)

Figure 3.8

Q1 = +100µC

8 cm Q3 =+300µC Q2 = -200µC

A

6 cm P X

Figure 3.9

3cm

4cm

3cm



6. Two point charges Q1 and Q2 are positioned at the corners of an equilateral

triangle as indicated in Figure 3.10.

(a) Find the resultant electric field at point A due to Q1 and Q2 charges

(b) If a charge of q = + 10 µ C is located at point A, what is the magnitude of

the force experienced by the charge?

APR 2008

(Ans : i) 95.24 x 106 N/C, θ = -19.18

o (4

th quarter @ below positive x-axis)

ii) 952.4 N

7. In Figure 3.11 shows the equipotential lines around a point charge.

(a) What is the work done to move a charge of 8.0µC from B to A?

(b) What is the work done to move a charge of 5.0 µC from B to C?

(c) (i) What is the potential difference between D and E?

(ii) What is the work done to move a charge of 2.0 µC from E to D

(iii) Estimate the average force on a charge of 2.0µC between E to D?

Figure 3.11

(Ans : a) -0.024 J V/m, b) 0 J, c) i) - 100 V ii) -2x10-4

J iii) 2.857x10-2

N)

Q2 Q1

Q1 = + 20 µ C

A

5 cm 5 cm

5 cm

Q2 = - 30 µ C

Figure 3.10



3.3 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 which a capacitor can store charge is called its capacitance. It is

defined as

V

QC =

where

C = 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 )

3.3.1 Charging and discharging a capacitor

Figure 3.12 shows an RC circuit that can be used to charge and discharge a capacitor.

Example 3.5

(a) If the charge on a capacitor is 50µC 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

C10200)V100)(F102(CVQ)b(

C102V25

C1050

V

QC)a(

66

66

−−

−−

×=×==

×=×

==

Figure 1.18

a S

b R

C

εεεε

Figure 3.12



Charging a capacitor

When switch S is first closed on a, there is no charge on the capacitor. Therefore, there is no

potential difference between the plates. Thus the full potential difference, ε appears across the

resistor, setting up an initial current, ε / R.

Once the current starts, charges began to appear on the capacitor plates and a potential

difference, Q / C builds up between those plates. This results in the potential difference across

the resistor to decrease by the same amount. This in turn reduces the charging current.

Consequently, the charge on the capacitor plates builds up and the charging current falls off

until the capacitor is fully charged.

Discharging a capacitor

Assume the capacitor in Figure 3.22 is fully charged to the voltage of the emf, ε. Switch S is

then moved from a to b. Since now the battery is disconnected, current will flow from the

capacitor.

Electrons flow from the negative plate to the positive plate of the capacitor, causing the net

charges on both plates to decrease, thus reducing the potential difference. The flow continues

until the net charges on the positive and negative plate is zero, resulting in zero potential

difference across the capacitor plates. The capacitor is now completely discharged.

The varying amount of charge in a capacitor during charging and discharging is illustrated in

Figure 3.13.

Parallel plate capacitor

Figure 3.14 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. We will also assume that the dielectric

material between the plates is free space / air / vacuum having the permittivity constant, ε0.

Q

Q0

Charging( Q = Cεεεε( 1 − − − − e-t/RC ) )

Discharging( Q = Qoe-t/RC )

Figure 1.19Figure 3.13



From the Gauss Theorem, the electric field strength between the plates is given by

A

QE

00 ε=

ε

σ= (a)

where

A

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

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

V

EA

V

QC 0ε

==

But

d

VE =

By substituting the equation, the capacitance of a parallel plate capacitor is then

d

AC 0ε

=

Equation above indicates that the capacitance of a parallel plate capacitor depends only on its

plate area and plate separation.

Area A

d

+ q

- q




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

pF550

F105.5m104.2

)m122.0)(m/F1085.8(

d

AC)a( 10

4

120

=

×=×

×=

ε= −

−

−

nC25

C1025)V45)(F10550(CVQ)b( 912

=

×=×== −−

Example 3.7

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

( )( )

( )( )

.C4.35 - of charge a has plate other the and

C4.35 of charge a has potential higher at plate The

C4.35V1010F103.54 CVQ(b)

3.54nF103.54

m1000.5

m00.2m/F1085.8

d

AC)a(

39-

9-

3

2120

µ

µ+

µ=××==

=×=

×

×=

ε=

−

−

( )( )

m/V1000.2m1000.5

V1010

d

VE

or

C/N1000.2m00.2F1085.8

C104.35

A

QE)c(

6

3

3

6

12

6

00

×=×

×==

×=×

×=

ε=

ε

σ=

−

−

−



3.3.2 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 1 shows the value of K for several dielectric materials.

Table 1 Values of K and dielectric strength for several dielectric material

Material Dielectric constant, K Dielectric strength(kV/mm)

Vacuum 1

Air (STP) 1.000576 3

Polystyrene 2.6 24

Ebonite 3

Paper 3.5 16

Pyrex 4.7 14

Mica 7

Water 80

The dielectric constant of a dielectric material is defined by

0C

CK =

where

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

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

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

factor of K, as in equation.

KV

V0 =



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 E0 is the

vacuum value and E the value with the dielectric, then

KE

E0 =

The capacitance of any capacitor is proportional to the permittivity of the material between its

plates. Therefore

00C

C

ε

ε=

where

ε = The permittivity of the dielectric material

ε0 = The permittivity of free space/ air / vacuum

By comparing equations,

0Kε=ε

In terms of ε, we can express the electric field within the dielectric as

ε

σ=E

The capacitance when the dielectric is present is given by

d

A

d

AKKCC 00 ε=ε==



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 0.2m2 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

( )( )

F107.17

m1000.1

m100.2m/F1085.8

d

AC)a(

11-

2

21120

0

×=

×

××=

ε=

−

−−

( )( )

( )( )

( )( )m/V1000.1

m100.2m/F106.26

C10531.0

A

QE

or

m/V1000.1m1000.1

V1000

d

VE)g(

m/V1000.3m1000.1

V3000

d

VE)f(

m/F106.263m/F1085.8K)e(

00.3F1017.7

F101.53

C

CK)d(

F101.53V1000.1

C10531.0

V

QC(c)

C10x531.0V103F1017.7 VCQ(b)

5

2112

60

5

2

5

2

0

12120

11-

11

0

11

3

60

63-11000

×=××

×=

ε=

×=×

==

×=×

==

×=×=ε=ε

=×

×==

×=×

×==

=××==

−−

−

−

−

−−

−

−−

−



Example 3.9

The electric field between the plates of a paper separated (K = 3.75) parallel plate

capacitor is 9.21 x 104V/m. The plates are 1.95mm apart and the charge on each plate is

0.775µC. Determine

(a) the capacitance of the capacitor.

(b) the area of each plate.

Solution

( )( )

( )( )( )

2

12

39

0

0

96

34

m254.01085.875.3

1095.11032.4

K

Cd A

d

AKC)b(

F1032.45.179

10775.0

V

QC

V5.1791095.1(1021.9EdV)a(

=×

××=

ε=

ε=

×=×

==

=××==

−

−−

−−

−

Example 3.10

A 5µF capacitor with air between the metal plates is connected to a 30V battery. The

battery is then removed, leaving the capacitor charged.

(a) Calculate the charge on the capacitor.

(b) The air between the plates is replaced by oil with K = 2.1. Find the new value of the

capacitance, the new potential difference in the capacitor.

Solution

(a) Q = CV = (5µ)(30) = 150µC

(b) The charges on the plates remain the same when the oil replaces the air since there

is no battery connected to the capacitor.

The capacitance increases by the factor of K

⇒C’ = KC (2.1)(5µ) = 10.5µF

The potential difference decreases by the factor of K

⇒V’=V/K = (30)/(2.1) = 14.3V



3.3.3 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 3.15a). 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 3.15b).

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

Since V = Q/C,

Q/Ceq = Q/C1 + Q/C2 + Q/C3

⇒1/Ceq = 1/C1 + 1/C2 + 1/C3

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,

∑=

=n

1i iS C

1

C

1

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

V1 = V2 = V3 = V

The total charges is the sum of the charges of individual capacitors,

Qeq = Q1 + Q2 + Q3

(a) (b)

Figure 3.15

C1 C2 C3

C1

C2

C3



Since Q = CV,

CeqV = C1V + C2V + C3V

Ceq = C1 + C2 + C3 (2.23)

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,

∑=

=n

1iiP CC (2.24)

Example 3.11

\Find the equivalent capacitance of the combination shown in Figure 3.16.

Solution

F18C C CC

other, each to parallel are C and C ,C

F4C

12

3

6

1

12

1

C

1

C

1

C

1

other, each to series in are C and C

21341234

2134

34

4334

43

µ=++=⇒

µ=⇒

µ=

µ+

µ=+=⇒

F6C

18

3

9

1

18

1

C

1

C

1

C

1

eq

51234eq

µ=⇒

µ=

µ+

µ=+=⇒

3µF

9µF

6µF

12µF

11µF

C1

C5

C4

C3

C2




Example 3.12

Calculate the charges on the capacitors shown in Figure 3.17 and the potential difference

across it.

Solution

F2C

6

3

6

1

3

1

C

1

C

1

C

1

F6C C Cother each to parallel in are C and C

T

YZXT

ZYYZZY

µ=⇒

µ=

µ+

µ=+=⇒

µ=+=⇒

For the circuit as a whole, the total charge is

( )( )

C240QQ series, in are C and C Since

C240120102VCQ

YZXYZX

6TTT

µ==

µ=×== −

For X,

40VV V,C to parallel is C Since

V40V80V120VVV

VV VSince

V803

240

C

QV

VCQ

,VisXacrosspotentialtheIf

ZYZY

XSYZ

YZXS

X

XX

XXX

X

==

=−=−=⇒

+=

=µ

µ==⇒

=

Figure 3.17

ε = 120V

2µF

4µF3µF

X

Y

Z

( )( )

( )( ) C16040104VCQ

, Vis Z across difference potential the If

Z, For

C8040102VCQ

, Vis Y across difference potential the If

Y,For

6ZZZ

Z

6YYY

Y

µ=×==⇒

µ=×==⇒

−

−



3.3.4 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 that is done in charging the capacitor is stored in the form of electrical potential

energy. It can be shown that the electrical potential energy stored in a charged capacitor is

C2

QU

2

=

By using equation Q=CV, it can be shown that U is also given by

QV2

1CV

2

1U 2 ==

Example 3.13

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, how

much additional charges will flow onto the capacitor from the 500V source?

Solution

( )( )

( )( )

( )( )

( )( )

nC35QQQ

is capacitor the onto flowing charge of amount the result, a As

nC57102260.2KQQTherefore,

before. than larger times 2.60 ecapacitanc a havenow willcapacitor The)c(

J5.550010222

1QV

2

1U

nC225001044CVQ)b(

pF44004.0

02.01085.8

d

AC)a(

0'

90

'

9

12

120

=−=∆

=×==

µ=×==

=×==

=×

=ε

=

−

−

−

−



Example 3.14

A 5µF capacitor(X) is fully charged by a 40V supply. The supply is then disconnected

the capacitor. It is then connected across an uncharged 20µF capacitor(Y) (See Figure

3.18). 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)(5x10-6

) = 200µC

When X and Y are connected, they are parallel to each other

⇒ Total capacitance, CT = CX + CY = 5µF + 20µF = 25µF

The total charge is unchanged, Q0 = QX + QY

⇒ Q0 = CTVT

200x10-6

C = 25x10-6

VT

VT = 200/25

= 8V

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

For Y,

QX = CXVT = 8 x (20x10-6

) = 160x10-6

C

Since the total charge is conserved,

⇒ Q0 = QX + QY

⇒ QX = Q0 - QY = (200 – 160)x10-6C = 40x10-6C

(c) The initial energy, Ui = (1/2)CXV02

= (1/2)(5x10-6)(40)2 = 4mJ

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

2

= (1/2)(5µF + 20µF)(8)2 = 0.8mJ

X

Y




Tutorial 3.2

1. 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)

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

placed 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)

3. 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 (K=3.5)

and half is filled with air (See Figure 3.19). Calculate the capacitance of the

capacitor.

(Answer: 0.50 nF)

4. Calculate the equivalent capacitance in Figure 3.20a and 3.20b.

(Answer : (a) 6µµµµF, (b) 3.33µµµµF )

paper air Figure 3.19

3µF 6µF

4µF

12 V 6µF

4µF

5µF

1000 V

(a) (b)

Figure 3.20



5. In Figure 3.21, calculate

(a) the equivalent capacitance in the circuit.

(b) the voltage across the 10µF capacitor in the circuit.

(Answer: a) 4.84µµµµF b) 2.32 V)

6. Four capacitors are connected across terminals a and b as shown in Figure 3.22.

(a) Calculate the equivalent capacitance across ab

(b) If a steady voltage of 100V is connected across ab, find

(i) the charge on the 5µF capacitor.

(ii) the voltage across the 10µF capacitor.

(Answer : (a) 2µµµµF (b) (i) 200µµµµC (ii) 20V)

7. Compute the energy stored in a 60 pF capacitor

(a) when charged to a pd of 2 kV

(b) when the charge on the plate is 30 nC.

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

8. A parallel plate capacitor having area 40 cm2 and spacing of 1 mm is charged to a

potential difference of 600V. Find

(a) the capacitance

(b) the magnitude of charge on each plate

(c) the stored energy

(d) the electric field between the plate

(Answer: (a) 35 pF (b) 21 nC (c) 6.3 µµµµJ (d) 6.0x105 V/m)

10µF

30µF

30µF

12µF

12µF

12 V Figure 3.21

Figure 3.22 5µF

10µF

3µF

2µF a

b



9. 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 above 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)

10. With reference to the capacitive circuit in Figure 3.23, find:

(a) the total electrical energy stored in the 5µF and 4µF capacitors.

(b) the electric charge stored in the 3µF capacitor.

(Answer : (a) 160µµµµJ (b) 36µµµµC)

Figure 3.23

11. A parallel plate capacitor consists of plate of area 1.5 x 10-4

m2 and separated by 1.0

mm. The capacitor is connected to a 12 V battery.

(a) What is the charge on the plate?

(b) How much energy is stored in the capacitor?

APR 2009

(Answer : 1.6 x 10-11

C, 9.6 x 10-11

J)

12. Three capacitors are connected as shown in Figure 3.24.

A 12 V potential difference is applied to the terminals.

(a) What is the total capacitance across the terminals

(b) Find the charge on each capacitor

(c) Find the voltage across each capacitor

APR 2008

(Answer : a) 6 µµµµF b) 48 µµµµF , 24 µµµµF c) 8 V, 4 V)

Figure 3.24

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