EP025Example2

Parallel plate capacitor :

Example 1 :

The plates of a parallel-plate capacitor are 8.0 mm apart and each has an area of 4.0 cm2. The
plates are in vacuum. If the potential difference across the plates is 2.0 kV, determine (a) the
capacitance of the capacitor (b) the amount of charge on each plate (c) the electric field strength
was produced.

Solution :

(a) The capacitance of the capacitor :

Step Particular Working Sheet

1 Target of the question : C = ?

2 Description (figure) :

3 Given information : d = 8.0x10-3m, A = 4.0x10-4m2, V = 2x103V.

4 Hidden information : -

5 Related equation / C = εoA
formula : d

6 Information required : -

C= (8.85x10 −12 )(4.0x10 −4 )

7 Computation : 8.0x10 −3

= 4.43x10-13F

The capacitance of the capacitor, C = 44.3pF (For every 1V of

8 Answer : potential difference across the capacitor, it able to accumulate

44.3pC of charge at each plate).

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(b) The amount of charge on each plate :

Step Particular Working Sheet

1 Target of the question : Q = ?

2 Description (figure) :

3 Given information : V = 2x103V, C = 4.43x10-13F

4 Hidden information : -

5 Related equation / C= Q
formula : V

6 Information required : -

7 Computation : 4.43x10-13 = Q
2x10 3

Q = 8.86x10-10C

8 Answer : The amount of charge on each plate is 88.6nC (as many as
5.54x109 electrons are accumulated on the plate).

(c) The electric field strength was produced :

Step Particular Working Sheet

1 Target of the question : E = ?

2 Description (figure) :

3 Given information : d = 8.0x10-3m, V = 2x103V.

4 Hidden information : -

5 Related equation / E= V
formula : d

6 Information required : -

E= V
d
7 Computation :
2x10 3
E= 8.0x10 −3 = 250kVm-1

8 Answer : The electric field strength was produced, E = 250kVm-1

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Capacitor connected in series, parallel and in
combination :

Example 1 :

Two capacitors, C1 = 2 µF and C2 = 4 µF are connected in series to a 12 V battery. Calculate (a)
equivalent capacitance (b) charge on each capacitor (c) potential difference across each

capacitor.

(a) The equivalent capacitance :

Step Particular Working Sheet
C1 = 2µF C2 = 4µF
1 Target of the question : CT = ?

2 Description (figure) : Q1, V1 Q2, V2

VT = 12V

3 Given information : C1 = 2x10-6F, C2 = 4x10-6F, V = 12 V
4 Hidden information : ‘connected in series’.

5 Related equation / 1 =1+ 1
formula : CT C1 C2

6 Information required : -

1 =1+1
CT C1 C2

7 Computation : = 1 + 1
2x10 −6 4x10 −6
CT = 1.33x10-6F

8 Answer : The equivalent capacitance, CT = 1.33x10-6F

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(b) The charge on each capacitor :

Step Particular Q1 = ? Working Sheet
1 Target of the question : Q2 = ?
C1 = 2µF C2 = 4µF
2 Description (figure) : Q1, V1 Q2, V2

VT = 12V

3 Given information : C1 = 2x10-6F, C2 = 4x10-6F, V = 12 V, CT = 1.33x10-6F

4 Hidden information : ‘connected in series’.

5 Related equation / Q1 = Q2 = QT ..(i)
formula :

QT = ?

6 Information required : From equation :
Q = CV

Then, QT = CTVT ..(ii)
Substitute (ii) into (i) :

7 Computation : Q1 = Q2 = QT = CTVT
= (1.33x10-6)(12)
= 1.6x10-5C

8 Answer : The charge on each capacitor, Q1 and Q2 has the same
magnitude as QT = 1.6x10-5C.

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(c) The potential difference across each capacitor :

Step Particular V1 = ? Working Sheet
1 Target of the question : V2 = ? C1 = 2µF C2 = 4µF

2 Description (figure) : Q1, V1 Q2, V2

VT = 12V

3 Given information : C1 = 2x10-6F, C2 = 4x10-6F, V = 12 V, Q1 = Q2 = QT = 1.6x10-5C

4 Hidden information : ‘connected in series’.

5 Related equation / Based on equation :
formula : Q = CV
Q1 = C1V1
Q2 = C2V2

6 Information required : - -

Q1 = C1V1 Q2 = C2V2
1.6x10-5 = (2x10-6)V1 1.6x10-5 = (4x10-6)V2
7 Computation :

V1 = 8V V2 = 4V

The potential difference across each capacitor is V1 = 8V and

V2 = 4V respectively.

8 Answer : Check the answer :

VT = V1 + V2
=8+4

= 12V#

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C1, Q1, C2, Q2, Example 2 :
V1 V2
Four capacitors are connected as shown in figure.
C3, Q3, C4, Q4, Calculate (a) the equivalent capacitance between points
V3 V4 a and b (b) the charge on each capacitor if Vab =15.0 V.

(a) The equivalent capacitance between points a and b :

Step Particular Working Sheet

1 Target of the question : CT = ?

C1, Q1, C2, Q2,
V1 V2

2 Description (figure) :

C3, Q3, C4, Q4,
V3 V4

3 Given information : C1 = 15x10-6F, C2 = 3x10-6F, C3 = 6x10-6F,
C4 = 20x10-6F V = 15 V

C1 & C2 → Series connection.

4 Hidden information : C12 & C3 → Parallel connection.

C123 & C4 → Series connection.

5 Related equation / 1= 1 +1 ..(i)
formula : CT C123 C4

C123 = C12 + C3 ..(ii)
6 Information required : 1 = 1 + 1 ..(iii)

C12 C1 C2
From equation (iii) :

1 =1+1
C12 C1 C2

= 1 + 1
15x10 −6 3x10 −6
C12 = 2.5x10-6F

7 Computation : Substitute (iii) into (ii) :

C123 = C12 + C3
C123 = 2.5x10-6 + 6x10-6

= 8.5x10-6F

Substitute (ii) into (i) :

1 = 1 + 1
CT 8.5x10 −6 20x10 −6

CT = 5.96x10-6F

8 Answer : The equivalent capacitance between points a and b,
CT = 5.96x10-6F

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(b) The charge on each capacitor :

Step Particular Working Sheet

1 Target of the question : Q1 = ?, Q2 = ?, Q3 = ?, Q4 = ?

2 Description (figure) : C1, Q1, C2, Q2,
V1 V2

C3, Q3, C4, Q4,
V3 V4

3 Given information : C1 = 15x10-6F, C2 = 3x10-6F, C3 = 6x10-6F,
C4 = 20x10-6F, CT = 5.96x10-6F, VT = 15V

Q1 = Q2 = Q12 .. (i)

Q12 + Q3 = Q123 = Q4 ..(ii)

4 Hidden information : Q123 = Q4 = QT ..(iii)

VT = V4 + V123 ..(iv)

V123 = V3 = V1 + V2 ..(v)

Based on equation :

Q = CV, then

5 Related equation / QT = CTVT ..(vi)
formula : Q4 = C4V4 ..(vii)
Q3 = C3V3 ..(viii)

Q2 = C2V2 ..(ix)

Q1 = C1V1 ..(x)

6 Information required : -

From equation (vi) :
QT = CTVT = (5.96x10-6)(15) = 8.94x10-5C

Substitute (vi) into (iii), then Q4 = 8.94x10-5C.

From equation (vii) :

Q4 = C4V4
(8.94x10-5) = (20x10-6)V4
V4 = 4.47V

Substitute (vii) into (iv) :

VT = V4 + V123
15 = 4.47 + V123
V123 = 10.53V

7 Computation : Substitute (iv) into (v), then V3 = 10.53V

Substitute (v) into (viii),
Q3 = C3V3 = (6x10-6)(10.53) = 6.318x10-5C

Based on equation (ii)
Q12 + Q3 = Q4
Q12 + 6.318x10-5 = 8.94x10-5
Q12 = 2.622x10-5C

Substitute (ii) into (i),
Q1 = Q2 = Q12 = 2.622x10-5C

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The charge on each capacitor :
Q1 = 2.622x10-5C
Q2 = 2.622x10-5C
8 Answer : Q3 = 6.318x10-5C

Q4 = 8.94x10-5C

Charging and discharging process :

Example 1 :

A capacitor of 3 µF is charged using a battery of 1.5 V. The charged capacitor is then discharged
through 50 kΩ resistor. (a) Determine the time constant of the discharge circuit (b) Calculate the
time taken for the charge on the capacitor to decrease to 1/100 of its initial value.

Solution :

(a) The time constant of the discharge circuit :

Step Particular Working Sheet
1 Target of the question : τ=?
2 -
3 Description (figure) : C = 3x10-6F, V = 1.5V, R = 50x103Ω
4 Given information : ‘Discharged’
5 Hidden information :
6 Related equation / τ = RC

7 formula : -
Information required : τ = RC
8
Computation : = (50x103)(3x10-6)
= 0.15s.
Answer : The time constant of the discharge circuit, τ = 0.15s (the time
taken for the capacitor to remain 37% of its initial charge during
discharging process).

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(a) The time taken for the charge on the capacitor to decrease to (i) 1/100 of its initial value :

Step Particular Working Sheet
1 Target of the question :
2 t=?
3 Description (figure) :
4 Given information : -
5 Hidden information : C = 3x10-6F, V = 1.5V, R = 50x103Ω, τ = 0.15s
6 Related equation / 1. ‘Discharged’
2. Q = Qo/100
7 formula :
Information required : −t
8
Computation : Q = Qoe RC
-
Answer :
Qo = −t
100
Qo e RC

ln 1  = ln − t 
 100  0.15 
e

-4.605 = - t
0.15

t = 0.69s

The time taken for the capacitor to remain 10% of its initial

charge during discharging process, t = 0.69s.

Dielectrics :

Example 1 :

Consider a parallel plate capacitor of capacitance C = 110 µF, the plate area A = 0.15 m2 and
dielectric constant κ = 6.9. (a) Calculate the separation between the plates (b) Determine the
electric field strength formed if a 10 V potential difference is applied across the capacitor. (c)
State two ways that can increase the capacitance without increasing the area of the capacitor
plate.

Solution :

(a) The separation between the plates :

Step Particular Working Sheet
1 Target of the question :
2 d=?
3 Description (figure) :
4 Given information : -
5 Hidden information : C = 110x10-6F, A = 0.15m2, κ = 6.9.
6 Related equation / dielectric constant κ.
C = κεo A
7 formula :
Information required : d
8
Computation : -
C = κεo A
Answer :
d
110x10-6 = (6.9)(8.85x10 −12 )(0.15)

d
d = 8.33x10-8m
The separation between the plates, d = 883µm.

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(b) The electric field strength formed if a 10 V potential difference is applied across the capacitor :

Step Particular Working Sheet
1 Target of the question :
2 E=?
3 Description (figure) :
4 Given information : -
5 Hidden information : V = 10V, d = 8.33x10-8m
6 Related equation / dielectric constant κ.

7 formula : E= V
Information required : d
8
Computation : -

Answer : E= V
d

= 10
8.33x10 −8
= 120x106Vm-1

The electric field strength formed, E = 120MVm-1.

(c) Two ways that can increase the capacitance without increasing the area of the capacitor
plate :

Based on equation :

C = κεo A
d

Then, C∝ κ
d

Hence, the two ways can be used in order to increase the capacitance without increasing

the area of the capacitor plate :

1. Decrease the separation, d between the capacitor.
2. Use a dielectric material with greater κ.

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