Capacitors and Capacitance Practice Questions — JAMB UTME

Question 1Question →

A $6\text{ }\mu\text{F}$ capacitor is connected in series with a parallel arrangement of a $2\text{ }\mu\text{F}$ capacitor and a $1\text{ }\mu\text{F}$ capacitor. If the entire circuit is connected across a $30\text{ V}$ d.c. power supply, what is the electric charge stored on the $2\text{ }\mu\text{F}$ capacitor?

20\text{ }\mu\text{C}

30\text{ }\mu\text{C}

40\text{ }\mu\text{C}

60\text{ }\mu\text{C}

Show answer & explanation

Answer: $40\text{ }\mu\text{C}$

Answer

The charge stored on the

2\text{ }\mu\text{F}

capacitor is

40\text{ }\mu\text{C}

.

To find the charge on the

2\text{ }\mu\text{F}

capacitor, we first combine the parallel capacitors (

2\text{ }\mu\text{F} + 1\text{ }\mu\text{F} = 3\text{ }\mu\text{F}

). Next, we combine this in series with the

6\text{ }\mu\text{F}

capacitor to get an overall capacitance

C_{eq} = 2\text{ }\mu\text{F}

. The total charge drawn from the

30\text{ V}

supply is

Q = 2\text{ }\mu\text{F} \times 30\text{ V} = 60\text{ }\mu\text{C}

. This total charge enters the parallel combination, giving a potential drop across the parallel branch of

V_p = 60\text{ }\mu\text{C} / 3\text{ }\mu\text{F} = 20\text{ V}

. Therefore, the charge on the

2\text{ }\mu\text{F}

capacitor is

2\text{ }\mu\text{F} \times 20\text{ V} = 40\text{ }\mu\text{C}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance of the parallel section

C_{p} = 2\text{ }\mu\text{F} + 1\text{ }\mu\text{F} = 3\text{ }\mu\text{F}

Capacitors in parallel add directly.

2

Calculate the total equivalent capacitance of the network

C_{eq} = \frac{6 \times 3}{6 + 3} = \frac{18}{9} = 2\text{ }\mu\text{F}

The

6\text{ }\mu\text{F}

capacitor and the

3\text{ }\mu\text{F}

parallel equivalent are in series.

3

Find the total charge supplied by the

30\text{ V}

source

Q_{total} = C_{eq} \times V = 2\text{ }\mu\text{F} \times 30\text{ V} = 60\text{ }\mu\text{C}

Total charge is the product of equivalent capacitance and total voltage.

4

Determine the potential difference across the parallel branch

V_{p} = \frac{Q_{total}}{C_{p}} = \frac{60\text{ }\mu\text{C}}{3\text{ }\mu\text{F}} = 20\text{ V}

The total charge flows through the series combination, creating a potential drop across the parallel combination equal to

Q_{total} / C_{p}

.

5

Calculate the charge on the

2\text{ }\mu\text{F}

capacitor

Q_{2\mu\text{F}} = C_{2\mu\text{F}} \times V_{p} = 2\text{ }\mu\text{F} \times 20\text{ V} = 40\text{ }\mu\text{C}

The charge on a specific capacitor in parallel is the product of its capacitance and the voltage across the parallel branch.

Key Concept

Charge and voltage distribution in mixed capacitor networks

Question 2Question →

Two $12\text{ }\mu\text{F}$ capacitors are connected in series, and this combination is placed in parallel with a third capacitor of capacitance $6\text{ }\mu\text{F}$ . What is the total equivalent capacitance of the arrangement?

3.0\text{ }\mu\text{F}

4.8\text{ }\mu\text{F}

12.0\text{ }\mu\text{F}

30.0\text{ }\mu\text{F}

Show answer & explanation

Answer: $12.0\text{ }\mu\text{F}$

Answer

The total equivalent capacitance of the network is

12.0\text{ }\mu\text{F}

.

Combining the two identical

12\text{ }\mu\text{F}

capacitors in series gives an equivalent capacitance of

6.0\text{ }\mu\text{F}

. Adding this in parallel with the

6.0\text{ }\mu\text{F}

capacitor yields

6.0\text{ }\mu\text{F} + 6.0\text{ }\mu\text{F} = 12.0\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance (

C_s

) of the two

12\text{ }\mu\text{F}

capacitors connected in series.

C_s = \frac{12 \times 12}{12 + 12} = 6.0\text{ }\mu\text{F}

Capacitors in series combine reciprocally:

\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}

.

2

Calculate the total equivalent capacitance (

C_{eq}

) by adding

C_s

to the parallel

6.0\text{ }\mu\text{F}

capacitor.

C_{eq} = C_s + C_3 = 6.0\text{ }\mu\text{F} + 6.0\text{ }\mu\text{F} = 12.0\text{ }\mu\text{F}

Capacitors in parallel add directly:

C_{eq} = C_{series} + C_{parallel}

.

Key Concept

Equivalent capacitance of series-parallel capacitor networks

Question 3Question →

A $3\text{ }\mu\text{F}$ capacitor and a $6\text{ }\mu\text{F}$ capacitor are connected in series across a direct current voltage source. What is the total equivalent capacitance of the combination, in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 2

Answer

The total equivalent capacitance of the combination is

2\text{ }\mu\text{F}

.

For capacitors connected in series, the reciprocal of the total equivalent capacitance is equal to the sum of the reciprocals of the individual capacitances. Substituting

3\text{ }\mu\text{F}

and

6\text{ }\mu\text{F}

gives

\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}\text{ }\mu\text{F}^{-1}

, which yields an equivalent capacitance of

2\text{ }\mu\text{F}

.

Step-by-Step Solution

1

State the formula for equivalent capacitance of two capacitors in series

\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}

Capacitors connected in series combine reciprocally, unlike resistors connected in series.

2

Substitute the values of

C_1

and

C_2

\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}\text{ }\mu\text{F}^{-1}

Find a common denominator and add the fractions.

3

Calculate the reciprocal to determine

C_{eq}

C_{eq} = 2\text{ }\mu\text{F}

Inverting

\frac{1}{2}

yields the total equivalent capacitance.

Key Concept

Equivalent Capacitance in Series

Question 4Question →

Two capacitors with capacitances of $3\text{ }\mu\text{F}$ and $6\text{ }\mu\text{F}$ are connected in parallel. This combination is then connected in series with a $9\text{ }\mu\text{F}$ capacitor across a $12\text{ }\text{V}$ direct current source. What is the total electrical energy stored in the network?

3.24 \times 10^{-4}\text{ }\text{J}

6.48 \times 10^{-4}\text{ }\text{J}

7.92 \times 10^{-4}\text{ }\text{J}

1.30 \times 10^{-3}\text{ }\text{J}

Show answer & explanation

Answer: $3.24 \times 10^{-4}\text{ }\text{J}$

Answer

The total electrical energy stored in the network is

3.24 \times 10^{-4}\text{ }\text{J}

.

First, find the equivalent capacitance of the parallel branch (

3\text{ }\mu\text{F} + 6\text{ }\mu\text{F} = 9\text{ }\mu\text{F}

). Next, combine this with the

9\text{ }\mu\text{F}

capacitor in series, yielding

C_{eq} = \frac{9 \times 9}{9 + 9} = 4.5\text{ }\mu\text{F} = 4.5 \times 10^{-6}\text{ }\text{F}

. Finally, substituting into the stored energy formula

E = \frac{1}{2} C_{eq} V^2

gives

E = 0.5 \times 4.5 \times 10^{-6} \times 144 = 3.24 \times 10^{-4}\text{ }\text{J}

.

Step-by-Step Solution

1

Calculate equivalent capacitance of the parallel section

C_p = C_1 + C_2 = 3\text{ }\mu\text{F} + 6\text{ }\mu\text{F} = 9\text{ }\mu\text{F}

Capacitors in parallel add directly.

2

Calculate total equivalent capacitance of the network

\frac{1}{C_{eq}} = \frac{1}{C_p} + \frac{1}{C_3} = \frac{1}{9\text{ }\mu\text{F}} + \frac{1}{9\text{ }\mu\text{F}} = \frac{2}{9\text{ }\mu\text{F}} \implies C_{eq} = 4.5\text{ }\mu\text{F} = 4.5 \times 10^{-6}\text{ }\text{F}

Capacitors in series combine via reciprocals.

3

Calculate total stored electrical energy

E = \frac{1}{2} C_{eq} V^2 = \frac{1}{2} \times (4.5 \times 10^{-6}\text{ }\text{F}) \times (12\text{ }\text{V})^2 = 3.24 \times 10^{-4}\text{ }\text{J}

Energy stored in a capacitor network depends on equivalent capacitance and potential difference across it.

Key Concept

Equivalent capacitance of series-parallel combinations and energy stored in a capacitor

Question 5Question →

A parallel plate capacitor with a capacitance of $12\text{ }\mu\text{F}$ is fully charged by connecting it across a $50\text{ V}$ direct current power source. What is the total electrostatic energy stored in the capacitor in millijoules ( $\text{mJ}$ )?

Show answer & explanation

Answer: 15

Answer

The total electrostatic energy stored in the capacitor is

15\text{ mJ}

.

The energy stored in a capacitor is given by

E = \frac{1}{2}CV^2

. Substituting

C = 12 \times 10^{-6}\text{ F}

and

V = 50\text{ V}

yields

E = \frac{1}{2} \times 12 \times 10^{-6} \times 2500 = 0.015\text{ J} = 15\text{ mJ}

.

Step-by-Step Solution

1

Convert given values to standard SI units.

C = 12 \times 10^{-6}\text{ F}

,

V = 50\text{ V}

Calculating in SI units ensures the resultant energy is in Joules.

2

Apply the formula for energy stored in a charged capacitor.

E = \frac{1}{2} C V^2

Work done during charging is stored as electrostatic potential energy in the electric field between the plates.

3

Substitute the values and convert Joules to millijoules.

E = \frac{1}{2} \times (12 \times 10^{-6}) \times 2500 = 0.015\text{ J} = 15\text{ mJ}

Multiplying Joules by

10^3

converts the result into millijoules.

Key Concept

Energy stored in a capacitor (

E = \frac{1}{2}CV^2

)

Question 6Question →

A capacitor of capacitance $5\text{ }\mu\text{F}$ is charged to a potential difference of $200\text{ V}$ and then disconnected from the power supply. If it is subsequently connected in parallel across an uncharged capacitor of capacitance $15\text{ }\mu\text{F}$ , what is the final common potential difference across the combination in volts?

Show answer & explanation

Answer: 50

Answer

The final common potential difference across the combination is 50 V.

By charge conservation, the total charge

Q = C_1 V_1 = 5\text{ }\mu\text{F} \times 200\text{ V} = 1000\text{ }\mu\text{C}

is shared across the two parallel capacitors. The total equivalent capacitance is

C_{\text{total}} = C_1 + C_2 = 20\text{ }\mu\text{F}

. Therefore, the final potential difference is

V = \frac{Q}{C_{\text{total}}} = \frac{1000\text{ }\mu\text{C}}{20\text{ }\mu\text{F}} = 50\text{ V}

.

Step-by-Step Solution

1

Calculate the initial electric charge (

Q

) stored on the charged capacitor

Q = 5\text{ }\mu\text{F} \times 200\text{ V} = 1000\text{ }\mu\text{C}

Before connection, all charge is stored solely on the

5\text{ }\mu\text{F}

capacitor.

2

Calculate the total equivalent capacitance (

C_{\text{total}}

) of the parallel network

C_{\text{total}} = 5\text{ }\mu\text{F} + 15\text{ }\mu\text{F} = 20\text{ }\mu\text{F}

Capacitors in parallel add directly (

C_{\text{total}} = C_1 + C_2

).

3

Apply the law of conservation of charge to find the final common voltage (

V

)

V = \frac{Q}{C_{\text{total}}} = \frac{1000\text{ }\mu\text{C}}{20\text{ }\mu\text{F}} = 50\text{ V}

The total charge remains conserved and redistributes across the total combined capacitance.

Key Concept

Charge Redistribution and Conservation in Parallel Capacitors

Question 7Question →

Two capacitors with capacitances of $10\text{ }\mu\text{F}$ and $15\text{ }\mu\text{F}$ are connected in parallel. What is the equivalent capacitance of the combination in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 25

Answer

The equivalent capacitance of the parallel combination is

25\text{ }\mu\text{F}

.

When capacitors are connected in parallel, the total equivalent capacitance is equal to the direct sum of the individual capacitances:

C_{\text{eq}} = C_1 + C_2 = 10\text{ }\mu\text{F} + 15\text{ }\mu\text{F} = 25\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Identify the relationship for parallel capacitors

C_{\text{eq}} = C_1 + C_2

Capacitors connected in parallel store charge independently across the same potential difference, so their capacitances add directly.

2

Substitute the given values into the formula

C_{\text{eq}} = 10\text{ }\mu\text{F} + 15\text{ }\mu\text{F}

The circuit contains two capacitors of

10\text{ }\mu\text{F}

and

15\text{ }\mu\text{F}

in parallel.

3

Calculate the total capacitance

25\text{ }\mu\text{F}

Simple addition of the two values yields

25\text{ }\mu\text{F}

.

Key Concept

Equivalent Capacitance of Parallel Connected Capacitors

Question 8Question →

Two capacitors of capacitances $C_1 = 6\text{ }\mu\text{F}$ and $C_2 = 12\text{ }\mu\text{F}$ are connected in series across a $180\text{ V}$ direct current power supply. After the capacitors are fully charged, the supply is disconnected. A dielectric material of dielectric constant $K = 4$ is then inserted to completely fill the space between the plates of $C_1$ . What is the new potential difference across $C_1$ ?

30\text{ V}

60\text{ V}

135\text{ V}

15\text{ V}

Show answer & explanation

Answer: $30\text{ V}$

Answer

30\text{ V}

The initial equivalent capacitance of the series combination is

4\text{ }\mu\text{F}

, which charges each capacitor to

720\text{ }\mu\text{C}

. Because the circuit is disconnected from the battery, charge is conserved. Inserting a dielectric of constant

K = 4

increases

C_1

to

24\text{ }\mu\text{F}

, resulting in a potential difference of

V = \frac{Q}{C_1'} = \frac{720\text{ }\mu\text{C}}{24\text{ }\mu\text{F}} = 30\text{ V}

.

Step-by-Step Solution

1

Calculate the initial equivalent capacitance of the series network

C_{eq} = 4\text{ }\mu\text{F}

Capacitors in series combine reciprocally:

\frac{1}{C_{eq}} = \frac{1}{6} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}\text{ }\mu\text{F}^{-1}

.

2

Determine the charge stored on each capacitor prior to disconnection

Q = 720\text{ }\mu\text{C}

Total charge supplied by the

180\text{ V}

battery is

Q = C_{eq} V = 4\text{ }\mu\text{F} \times 180\text{ V} = 720\text{ }\mu\text{C}

. In series, each capacitor holds this same charge.

3

Calculate the modified capacitance of

C_1

after dielectric insertion

C_1' = 24\text{ }\mu\text{F}

A dielectric of constant

K = 4

scales capacitance by

K

:

C_1' = K \times C_1 = 4 \times 6\text{ }\mu\text{F} = 24\text{ }\mu\text{F}

.

4

Calculate the final potential difference across

C_1

using charge conservation

V_1' = 30\text{ V}

Because the source is disconnected, charge

Q = 720\text{ }\mu\text{C}

on

C_1

remains constant. Therefore,

V_1' = \frac{Q}{C_1'} = \frac{720\text{ }\mu\text{C}}{24\text{ }\mu\text{F}} = 30\text{ V}

.

Key Concept

Effect of dielectrics and charge conservation in disconnected series capacitor circuits

Question 9Question →

A parallel plate air capacitor of capacitance $8.0\text{ }\mu\text{F}$ is fully charged using a $40.0\text{ V}$ d.c. power supply and then disconnected from the source. A dielectric slab with a relative permittivity of $4.0$ is subsequently inserted to completely fill the region between the plates. Calculate the magnitude of the decrease in electrostatic energy stored in the capacitor, in microjoules ( $\mu\text{J}$ ).

Show answer & explanation

Answer: 4800

Answer

The magnitude of the decrease in stored electrostatic energy is 4800 μJ.

Disconnecting the battery ensures that the charge

Q = C_0 V = 320\text{ }\mu\text{C}

on the plates stays fixed. Inserting the dielectric increases capacitance fourfold to

32.0\text{ }\mu\text{F}

. The energy decreases from

U_i = 6400\text{ }\mu\text{J}

to

U_f = \frac{Q^2}{2C_f} = 1600\text{ }\mu\text{J}

, giving a total decrease of

4800\text{ }\mu\text{J}

.

Step-by-Step Solution

1

Calculate the initial energy stored in the air capacitor before disconnection.

Initial energy

U_i = \frac{1}{2} C_0 V^2 = \frac{1}{2} \times 8.0 \times 10^{-6} \text{ F} \times (40.0 \text{ V})^2 = 6.4 \times 10^{-3} \text{ J} = 6400 \text{ } \mu\text{J}

.

The initial state has known capacitance and potential difference.

2

Calculate the new capacitance with the dielectric present.

Final capacitance

C_f = K C_0 = 4.0 \times 8.0 \text{ } \mu\text{F} = 32.0 \text{ } \mu\text{F}

.

Inserting a dielectric of constant

K

scales the capacitance by

K

.

3

Calculate the final stored energy using charge conservation.

Final energy

U_f = \frac{U_i}{K} = \frac{6400 \text{ } \mu\text{J}}{4.0} = 1600 \text{ } \mu\text{J}

.

Disconnection forces the charge

Q

to remain fixed, so energy scales inversely with capacitance (

U = \frac{Q^2}{2C}

).

4

Find the difference between initial and final energy.

\Delta U = 6400 \text{ } \mu\text{J} - 1600 \text{ } \mu\text{J} = 4800 \text{ } \mu\text{J}.

The loss in electrostatic energy represents the work done by the field pulling the dielectric slab into the plates.

Key Concept

Effect of dielectric insertion on stored electrostatic energy under isolated (constant charge) conditions

Estimated Time:2m 0s

Question 10Question →

A parallel-plate capacitor with air between its plates has a capacitance of $12\text{ }\mu\text{F}$ . A dielectric slab of relative permittivity $\varepsilon_r = 4$ and thickness $t = \frac{d}{3}$ , where $d$ is the total plate separation, is inserted between the plates parallel to them. What is the effective capacitance of the modified capacitor?

16\text{ }\mu\text{F}

48\text{ }\mu\text{F}

162\text{ }\mu\text{F}

6\text{ }\mu\text{F}

Show answer & explanation

Answer: $16\text{ }\mu\text{F}$

Answer

The effective capacitance of the modified capacitor is

16\text{ }\mu\text{F}

.

Inserting a dielectric slab of thickness

t = d/3

creates a system equivalent to two series capacitors: an air-filled region of thickness

2d/3

(

C_1 = 1.5 C_0 = 18\text{ }\mu\text{F}

) and a dielectric-filled region of thickness

d/3

(

C_2 = 12 C_0 = 144\text{ }\mu\text{F}

). Combining them via the series reciprocal formula gives

C_{\text{eq}} = \frac{18 \times 144}{18 + 144} = 16\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Model the partially filled capacitor as two capacitors connected in series.

Air layer of thickness

d_1 = d - t = \frac{2}{3}d

forms capacitor

C_1

. Dielectric layer of thickness

d_2 = t = \frac{1}{3}d

forms capacitor

C_2

.

Dividing the plate gap vertically into two distinct media creates two capacitive regions sharing the same electric flux path.

2

Calculate the individual capacitances

C_1

and

C_2

in terms of initial air capacitance

C_0 = 12\text{ }\mu\text{F}

.

C_1 = \frac{\varepsilon_0 A}{\frac{2}{3}d} = \frac{3}{2}C_0 = \frac{3}{2}(12) = 18\text{ }\mu\text{F}

and

C_2 = \frac{\varepsilon_r \varepsilon_0 A}{\frac{1}{3}d} = 3 \varepsilon_r C_0 = 3(4)(12) = 144\text{ }\mu\text{F}

.

Capacitance is inversely proportional to plate distance and directly proportional to relative permittivity.

3

Calculate the equivalent capacitance

C_{\text{eq}}

for two series capacitors.

\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{18} + \frac{1}{144} = \frac{8 + 1}{144} = \frac{9}{144} = \frac{1}{16}\text{ }\mu\text{F}^{-1} \implies C_{\text{eq}} = 16\text{ }\mu\text{F}

.

Capacitors connected in series combine reciprocally.

Key Concept

Partially filled parallel-plate capacitors act as series combinations of distinct capacitive layers.

Estimated Time:3m 0s

Question 11Question →

Two capacitors with capacitances of $3.0\text{ }\mu\text{F}$ and $6.0\text{ }\mu\text{F}$ are connected in parallel across a direct-current source. What is the equivalent capacitance of this combination?

9.0\text{ }\mu\text{F}

2.0\text{ }\mu\text{F}

4.5\text{ }\mu\text{F}

18.0\text{ }\mu\text{F}

Show answer & explanation

Answer: $9.0\text{ }\mu\text{F}$

Answer

The equivalent capacitance of the parallel combination is

9.0\text{ }\mu\text{F}

.

When capacitors are connected in parallel, each capacitor experiences the full potential difference of the voltage source, and the total charge stored is the sum of individual charges (

Q_{total} = Q_1 + Q_2

). Thus, the equivalent capacitance is the direct sum of the individual capacitances:

C_{eq} = 3.0\text{ }\mu\text{F} + 6.0\text{ }\mu\text{F} = 9.0\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Identify the combination rule for parallel capacitors.

The total capacitance is given by

C_{eq} = C_1 + C_2

.

Capacitors in parallel share the same potential difference, so total charge stored is the sum of individual charges.

2

Substitute the given values into the parallel capacitance formula.

C_{eq} = 3.0\text{ }\mu\text{F} + 6.0\text{ }\mu\text{F} = 9.0\text{ }\mu\text{F}

.

Direct addition yields the total equivalent capacitance.

Key Concept

Parallel Combination of Capacitors

Estimated Time:45s

Question 12Question →

Two identical air-filled parallel-plate capacitors, $C_1$ and $C_2$ , each of capacitance $C$ , are connected in series across a direct-current voltage source of potential difference $V$ . While the circuit remains connected to the voltage source, a dielectric slab of relative permittivity $\varepsilon_r = 3$ is fully inserted into $C_1$ , completely filling the space between its plates. What is the ratio of the electrostatic energy stored in $C_1$ after inserting the dielectric to the energy stored in $C_1$ before the insertion?

3 : 4

1 : 3

3 : 1

27 : 4

Show answer & explanation

Answer: $3 : 4$

Answer

The ratio of the energy stored in the first capacitor after dielectric insertion to before insertion is

3 : 4

(or

0.75

).

Initially, the two identical capacitors divide the total source voltage

V

equally, giving

V_1 = V/2

and initial energy

U_{1,i} = \frac{1}{8} C V^2

. When the dielectric of relative permittivity

3

is inserted, the capacitance of the first capacitor becomes

3C

. In a series circuit connected to a constant voltage source, the total charge becomes

Q = C_{eq}V = \frac{3}{4}CV

, which reduces the voltage across the modified capacitor to

V_1' = \frac{Q}{3C} = \frac{V}{4}

. The new stored energy is

U_{1,f} = \frac{1}{2}(3C)(\frac{V}{4})^2 = \frac{3}{32} C V^2

. Dividing

U_{1,f}

by

U_{1,i}

yields

\frac{3/32}{1/8} = \frac{3}{4}

.

Step-by-Step Solution

1

Calculate initial capacitance and voltage across

C_1

Initial capacitance

C_{1,i} = C

. Since

C_1

and

C_2

are identical and in series, initial potential difference across

C_1

is

V_{1,i} = \frac{V}{2}

.

Equal capacitors in series divide total voltage equally.

2

Calculate initial electrostatic energy stored in

C_1

U_{1,i} = \frac{1}{2} C_{1,i} V_{1,i}^2 = \frac{1}{2} C \left(\frac{V}{2}\right)^2 = \frac{1}{8} C V^2

.

Formula for energy stored in a capacitor is

U = \frac{1}{2} C V^2

.

3

Determine final capacitance of

C_1

and new voltage division

New capacitance

C_{1,f} = \varepsilon_r C = 3C

. Total equivalent capacitance

C_{eq} = \frac{3C \cdot C}{3C + C} = \frac{3}{4} C

. Total charge supplied

Q = C_{eq} V = \frac{3}{4} C V

. Final voltage across

C_1

is

V_{1,f} = \frac{Q}{C_{1,f}} = \frac{\frac{3}{4} C V}{3C} = \frac{V}{4}

.

Dielectric increases capacitance by factor

\varepsilon_r

, altering equivalent capacitance and potential distribution in series.

4

Calculate final electrostatic energy in

C_1

and compute the ratio

U_{1,f} = \frac{1}{2} (3C) \left(\frac{V}{4}\right)^2 = \frac{3}{32} C V^2

. Ratio

\frac{U_{1,f}}{U_{1,i}} = \frac{\frac{3}{32} C V^2}{\frac{1}{8} C V^2} = \frac{3}{4}

.

Divide final stored energy by initial stored energy.

Key Concept

Series combination of capacitors with dielectric insertion under constant battery voltage

Question 13Question →

Three capacitors, each of capacitance $12\text{ }\mu\text{F}$ , are arranged such that two of them are connected in parallel, and this combination is connected in series with the third capacitor. If the entire network is connected across a $20\text{ V}$ d.c. power supply, what is the total energy stored in the network?

0.8 \times 10^{-3}\text{ J}

1.6 \times 10^{-3}\text{ J}

3.6 \times 10^{-3}\text{ J}

7.2 \times 10^{-3}\text{ J}

Show answer & explanation

Answer: $1.6 \times 10^{-3}\text{ J}$

Answer

The total energy stored in the network is

1.6 \times 10^{-3}\text{ J}

.

Combining two

12\text{ }\mu\text{F}

capacitors in parallel gives a parallel capacitance of

24\text{ }\mu\text{F}

. Connecting this combination in series with the third

12\text{ }\mu\text{F}

capacitor results in an equivalent network capacitance of

C_{\text{eq}} = \frac{24 \times 12}{24 + 12} = 8\text{ }\mu\text{F}

. Substituting this into the energy formula

E = \frac{1}{2} C_{\text{eq}} V^2

yields

E = \frac{1}{2} \times 8 \times 10^{-6} \times 400 = 1.6 \times 10^{-3}\text{ J}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance of the parallel branch.

C_p = 12\text{ }\mu\text{F} + 12\text{ }\mu\text{F} = 24\text{ }\mu\text{F}

Capacitors in parallel add directly.

2

Calculate the total equivalent capacitance of the network.

C_{\text{eq}} = \frac{C_p \times C_3}{C_p + C_3} = \frac{24 \times 12}{24 + 12} = \frac{288}{36} = 8\text{ }\mu\text{F} = 8 \times 10^{-6}\text{ F}

The parallel combination is connected in series with the third capacitor.

3

Calculate the total electrical energy stored in the combination.

E = \frac{1}{2} C_{\text{eq}} V^2 = \frac{1}{2} \times (8 \times 10^{-6}\text{ F}) \times (20\text{ V})^2 = 4 \times 10^{-6} \times 400 = 1.6 \times 10^{-3}\text{ J}

The formula for energy stored in a capacitor network is

E = \frac{1}{2} C V^2

.

Key Concept

Mixed capacitor networks and energy storage

Estimated Time:1m 30s

Question 14Question →

In an electric circuit, two capacitors $C_1 = 12\text{ }\mu\text{F}$ and $C_2 = 6\text{ }\mu\text{F}$ are connected in series. This series combination is then connected in parallel with a third capacitor $C_3$ of unknown value. When a direct-current potential difference of $100\text{ V}$ is applied across the entire network, the total electrostatic energy stored in the circuit is $100\text{ mJ}$ . What is the capacitance of $C_3$ in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 16

Answer

The capacitance of

C_3

is

16\text{ }\mu\text{F}

.

First, the series combination of

12\text{ }\mu\text{F}

and

6\text{ }\mu\text{F}

yields an equivalent branch capacitance of

4\text{ }\mu\text{F}

. Second, using

E = \frac{1}{2} C_{eq} V^2

with

E = 0.100\text{ J}

and

V = 100\text{ V}

gives a total circuit equivalent capacitance of

20\text{ }\mu\text{F}

. Finally, subtracting the branch capacitance from the total parallel equivalent capacitance gives

C_3 = 20\text{ }\mu\text{F} - 4\text{ }\mu\text{F} = 16\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Calculate the effective capacitance of the series branch containing

C_1

and

C_2

C_{12} = 4\text{ }\mu\text{F}

Capacitors in series combine reciprocally:

\frac{1}{C_{12}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{12} + \frac{1}{6} = \frac{3}{12} \implies C_{12} = 4\text{ }\mu\text{F}

.

2

Determine the total equivalent capacitance

C_{eq}

of the circuit using the given stored energy and voltage

C_{eq} = 20\text{ }\mu\text{F}

Energy stored in a capacitor network is

E = \frac{1}{2} C_{eq} V^2

. Rearranging gives

C_{eq} = \frac{2E}{V^2} = \frac{2 \times 0.100\text{ J}}{(100\text{ V})^2} = 20 \times 10^{-6}\text{ F} = 20\text{ }\mu\text{F}

.

3

Calculate the unknown capacitance

C_3

from the parallel combination formula

C_3 = 16\text{ }\mu\text{F}

Because the branch

C_{12}

and

C_3

are in parallel,

C_{eq} = C_{12} + C_3 \implies 20\text{ }\mu\text{F} = 4\text{ }\mu\text{F} + C_3 \implies C_3 = 16\text{ }\mu\text{F}

.

Key Concept

Series and parallel combinations of capacitors combined with electrostatic energy storage

Estimated Time:2m 0s

Question 15Question →

A $5.0\text{ }\mu\text{F}$ parallel-plate capacitor is connected across a potential difference of $20.0\text{ V}$ . What is the magnitude of the electric charge stored on either plate of the capacitor, in microcoulombs ( $\mu\text{C}$ )?

Show answer & explanation

Answer: 100

Answer

The magnitude of the electric charge stored on either plate of the capacitor is

100.0\text{ }\mu\text{C}

.

Using the capacitor charge equation

Q = C \times V

, substituting

C = 5.0\text{ }\mu\text{F}

and

V = 20.0\text{ V}

yields

Q = 5.0 \times 20.0 = 100.0\text{ }\mu\text{C}

.

Step-by-Step Solution

1

Identify the given physical quantities and the formula for electric charge on a capacitor.

Capacitance

C = 5.0\text{ }\mu\text{F}

, voltage

V = 20.0\text{ V}

. Formula:

Q = C V

.

The charge stored by a capacitor is directly proportional to the potential difference across its terminals.

2

Perform the multiplication to determine the charge magnitude in microcoulombs.

Q = 5.0 \times 20.0 = 100.0\text{ }\mu\text{C}

.

Multiplying capacitance in microfarads by potential difference in volts yields charge directly in microcoulombs.

Key Concept

Fundamental relationship between capacitance, charge, and potential difference (

Q = C V

)

Estimated Time:45s

Question 16Question →

A parallel-plate capacitor with air between its plates has a capacitance of $15\text{ }\mu\text{F}$ . If the plate separation is reduced to one-third of its initial value and a dielectric material of relative permittivity $\varepsilon_r = 4.0$ is completely inserted between the plates, what is the new capacitance of the capacitor?

20\text{ }\mu\text{F}

45\text{ }\mu\text{F}

60\text{ }\mu\text{F}

180\text{ }\mu\text{F}

Show answer & explanation

Answer: $180\text{ }\mu\text{F}$

Answer

The new capacitance of the capacitor is

180\text{ }\mu\text{F}

.

The capacitance of a parallel-plate capacitor is given by

C = \frac{\varepsilon_r \varepsilon_0 A}{d}

. Reducing plate separation to one-third increases capacitance by a factor of 3. Adding a dielectric with relative permittivity

\varepsilon_r = 4.0

increases capacitance by a factor of 4. Combining both effects increases capacitance by a total factor of

3 \times 4 = 12

, yielding

12 \times 15\text{ }\mu\text{F} = 180\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Express the initial capacitance

C_1

using the formula for a parallel-plate air capacitor.

C_1 = \frac{\varepsilon_0 A}{d} = 15\text{ }\mu\text{F}

Air has a relative permittivity of 1.

2

Write the formula for the modified capacitance

C_2

with plate separation

d' = \frac{d}{3}

and dielectric constant

\varepsilon_r = 4.0

.

C_2 = \frac{\varepsilon_r \varepsilon_0 A}{d'} = \frac{4.0 \varepsilon_0 A}{\frac{d}{3}} = 4.0 \times 3 \times \left(\frac{\varepsilon_0 A}{d}\right) = 12 C_1

Reducing distance to

d/3

increases capacitance by a factor of 3, and inserting the dielectric increases capacitance by a factor of 4.

3

Calculate the value of the new capacitance.

C_2 = 12 \times 15\text{ }\mu\text{F} = 180\text{ }\mu\text{F}

Multiplying the combined scaling factor by the initial capacitance gives the final answer.

Key Concept

Parallel Plate Capacitance and Dielectrics

Estimated Time:1m 30s

Question 17Question →

A $2.0\text{ }\mu\text{F}$ capacitor and a $3.0\text{ }\mu\text{F}$ capacitor are connected in series across a $100\text{ V}$ d.c. power supply. What is the magnitude of the electric charge stored on the $2.0\text{ }\mu\text{F}$ capacitor?

500\text{ }\mu\text{C}

120\text{ }\mu\text{C}

200\text{ }\mu\text{C}

80\text{ }\mu\text{C}

Show answer & explanation

Answer: $120\text{ }\mu\text{C}$

Answer

The electric charge stored on the

2.0\text{ }\mu\text{F}

capacitor is

120\text{ }\mu\text{C}

.

The correct answer is

120\text{ }\mu\text{C}

. For capacitors connected in series, the equivalent capacitance

C_{eq}

is determined using

\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}

, giving

C_{eq} = \frac{2.0 \times 3.0}{2.0 + 3.0} = 1.2\text{ }\mu\text{F}

. Multiplying by the source voltage of

100\text{ V}

yields a total charge

Q = C_{eq} V = 120\text{ }\mu\text{C}

. Because capacitors in series store equal amounts of charge, the charge on the

2.0\text{ }\mu\text{F}

capacitor is

120\text{ }\mu\text{C}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance

C_{eq}

of the series combination.

C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{2.0 \times 3.0}{2.0 + 3.0} = 1.2\text{ }\mu\text{F}

Capacitors in series combine according to the reciprocal formula

\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}

.

2

Determine the total charge

Q

supplied by the

100\text{ V}

source.

Q = C_{eq} V = 1.2\text{ }\mu\text{F} \times 100\text{ V} = 120\text{ }\mu\text{C}

The total charge is the product of the equivalent capacitance and the total voltage.

3

Identify the charge on the individual

2.0\text{ }\mu\text{F}

capacitor.

Q_1 = Q = 120\text{ }\mu\text{C}

Components connected in series carry the exact same electric charge.

Key Concept

Equivalent Capacitance and Charge Distribution in Series Circuits

Estimated Time:1m 30s

Question 18Question →

Two capacitors with capacitances $C_1 = 4.0\text{ }\mu\text{F}$ and $C_2 = 12.0\text{ }\mu\text{F}$ are connected in series across a $120.0\text{ V}$ d.c. power supply. After becoming fully charged, the capacitors are disconnected from the supply and reconnected in parallel with plates of like polarity connected together. What is the total electrostatic energy lost in the reconnection process?

5.4 \times 10^{-3}\text{ J}

1.62 \times 10^{-2}\text{ J}

1.76 \times 10^{-2}\text{ J}

2.16 \times 10^{-2}\text{ J}

Show answer & explanation

Answer: $5.4 \times 10^{-3}\text{ J}$

Answer

The total electrostatic energy lost in the reconnection process is

5.4 \times 10^{-3}\text{ J}

.

The correct answer is derived by finding the total initial energy stored in series (

2.16 \times 10^{-2}\text{ J}

), determining the total combined charge (

720\text{ }\mu\text{C}

) and parallel capacitance (

16.0\text{ }\mu\text{F}

) upon reconnection to find the final stored energy (

1.62 \times 10^{-2}\text{ J}

), and calculating the difference of

5.4 \times 10^{-3}\text{ J}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance and charge of the initial series arrangement.

C_s = \frac{C_1 C_2}{C_1 + C_2} = \frac{4.0 \times 12.0}{4.0 + 12.0} = 3.0\text{ }\mu\text{F}

, so charge on each capacitor

Q = C_s V = (3.0 \times 10^{-6}\text{ F})(120\text{ V}) = 360\text{ }\mu\text{C}

.

Capacitors in series store identical charge equal to the product of equivalent series capacitance and total applied voltage.

2

Calculate the initial total electrostatic energy stored.

E_i = \frac{1}{2} C_s V^2 = \frac{1}{2} (3.0 \times 10^{-6}\text{ F})(120\text{ V})^2 = 2.16 \times 10^{-2}\text{ J}

.

Initial stored energy is determined by the series combination connected across the supply voltage.

3

Determine total charge and equivalent capacitance after parallel reconnection.

Q_p = Q_1 + Q_2 = 360\text{ }\mu\text{C} + 360\text{ }\mu\text{C} = 720\text{ }\mu\text{C}

and

C_p = C_1 + C_2 = 4.0\text{ }\mu\text{F} + 12.0\text{ }\mu\text{F} = 16.0\text{ }\mu\text{F}

.

Connecting like-polarity plates aggregates the individual charges and sums the capacitances in parallel.

4

Calculate the final potential difference and final stored energy.

V_p = \frac{Q_p}{C_p} = \frac{720\text{ }\mu\text{C}}{16.0\text{ }\mu\text{F}} = 45\text{ V}

, so

E_f = \frac{1}{2} C_p V_p^2 = \frac{1}{2} (16.0 \times 10^{-6}\text{ F})(45\text{ V})^2 = 1.62 \times 10^{-2}\text{ J}

.

Charge redistributes until both capacitors reach a common potential difference

V_p

.

5

Calculate the energy lost during reconnection.

\Delta E = E_i - E_f = 2.16 \times 10^{-2}\text{ J} - 1.62 \times 10^{-2}\text{ J} = 5.4 \times 10^{-3}\text{ J}

.

The energy dissipated as heat and spark during charge redistribution is the difference between initial and final total energies.

Key Concept

Energy dissipation during charge sharing between reconnected capacitors

Question 19Question →

A $6.0\text{ }\mu\text{F}$ capacitor is charged to a potential difference of $100\text{ V}$ using a direct-current source and then disconnected. It is subsequently connected in parallel across an uncharged $4.0\text{ }\mu\text{F}$ capacitor. What is the total electrostatic potential energy lost in the system during the redistribution of charge, in millijoules (mJ)?

Show answer & explanation

Answer: 12

Answer

The total electrostatic potential energy lost in the system during the redistribution of charge is 12 mJ.

The initial energy stored in the charged capacitor is

U_i = \frac{1}{2} C_1 V_1^2 = \frac{1}{2}(6.0 \times 10^{-6}\text{ F})(100\text{ V})^2 = 30\text{ mJ}

. When connected in parallel to the uncharged capacitor, the total charge

Q = 600\text{ }\mu\text{C}

is conserved across an equivalent capacitance of

C_{eq} = 6.0\text{ }\mu\text{F} + 4.0\text{ }\mu\text{F} = 10.0\text{ }\mu\text{F}

. The common potential becomes

V_f = \frac{Q}{C_{eq}} = 60\text{ V}

, leading to a final stored energy

U_f = \frac{1}{2} C_{eq} V_f^2 = 18\text{ mJ}

. The energy lost is

\Delta U = U_i - U_f = 30\text{ mJ} - 18\text{ mJ} = 12\text{ mJ}

.

Step-by-Step Solution

1

Calculate the initial stored charge

Q

and initial energy

U_i

in the charged

6.0\text{ }\mu\text{F}

capacitor.

Q = 6.0 \times 10^{-4}\text{ C} = 600\text{ }\mu\text{C}

and

U_i = 3.0 \times 10^{-2}\text{ J} = 30\text{ mJ}

.

Before connection, all charge and energy reside solely on the first capacitor.

2

Find the equivalent capacitance

C_{eq}

when the two capacitors are connected in parallel.

C_{eq} = 6.0\text{ }\mu\text{F} + 4.0\text{ }\mu\text{F} = 10.0\text{ }\mu\text{F}

.

Capacitances add directly when connected in parallel.

3

Determine the common final potential difference

V_f

across the combination.

V_f = \frac{Q}{C_{eq}} = \frac{600\text{ }\mu\text{C}}{10.0\text{ }\mu\text{F}} = 60\text{ V}

.

Total electric charge is conserved during redistribution between connected capacitors.

4

Calculate the final total energy

U_f

stored in the combined system.

U_f = \frac{1}{2} C_{eq} V_f^2 = \frac{1}{2} (10.0 \times 10^{-6}\text{ F})(60\text{ V})^2 = 18\text{ mJ}

.

Both capacitors now store energy under the new common potential difference.

5

Compute the total energy lost

\Delta U = U_i - U_f

.

\Delta U = 30\text{ mJ} - 18\text{ mJ} = 12\text{ mJ}

.

The difference in energy is dissipated as heat in connecting wires and spark/radiation.

Key Concept

Charge Conservation and Energy Dissipation during Charge Sharing in Capacitors

Estimated Time:2m 0s

Question 20Question →

Two capacitors of capacitances $8.0\text{ }\mu\text{F}$ and $4.0\text{ }\mu\text{F}$ are connected in parallel. This parallel combination is then connected in series with a single $4.0\text{ }\mu\text{F}$ capacitor across a $36.0\text{ V}$ d.c. voltage source. What is the potential difference across the single $4.0\text{ }\mu\text{F}$ capacitor?

9.0\text{ V}

14.4\text{ V}

27.0\text{ V}

36.0\text{ V}

Show answer & explanation

Answer: $27.0\text{ V}$

Answer

The potential difference across the single

4.0\text{ }\mu\text{F}

capacitor is

27.0\text{ V}

.

The two parallel capacitors (

8.0\text{ }\mu\text{F}

and

4.0\text{ }\mu\text{F}

) combine directly to give an equivalent capacitance of

12.0\text{ }\mu\text{F}

. This equivalent capacitor is in series with the single

4.0\text{ }\mu\text{F}

capacitor across the

36.0\text{ V}

source. Using the voltage divider rule for series capacitors, the voltage across the single

4.0\text{ }\mu\text{F}

capacitor is

V = 36.0 \times \frac{12.0}{4.0 + 12.0} = 27.0\text{ V}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance of the two parallel capacitors.

C_p = 8.0\text{ }\mu\text{F} + 4.0\text{ }\mu\text{F} = 12.0\text{ }\mu\text{F}

Capacitors in parallel add algebraically.

2

Calculate the total equivalent capacitance of the entire circuit.

C_T = \frac{4.0 \times 12.0}{4.0 + 12.0} = \frac{48.0}{16.0} = 3.0\text{ }\mu\text{F}

The single

4.0\text{ }\mu\text{F}

capacitor and the

12.0\text{ }\mu\text{F}

parallel combination are in series.

3

Find the total charge supplied by the battery.

Q = C_T \times V = 3.0\text{ }\mu\text{F} \times 36.0\text{ V} = 108.0\text{ }\mu\text{C}

The total charge is equal to the total capacitance multiplied by the total voltage.

4

Calculate the potential difference across the single

4.0\text{ }\mu\text{F}

capacitor.

V_1 = \frac{Q}{C_1} = \frac{108.0\text{ }\mu\text{C}}{4.0\text{ }\mu\text{F}} = 27.0\text{ V}

In a series connection, the charge on the single capacitor equals the total charge.

Key Concept

Potential difference distribution in mixed series-parallel capacitor networks

Estimated Time:1m 30s

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