Energy Stored in a Capacitor

Electrostatic Potential and Capacitance · CBSE

Theory

The derivation

Start with two uncharged conductors and move charge from conductor 2 to conductor 1 in small instalments. At the stage where the charges are $Q'$ and $-Q'$ , the potential difference is $V'=\dfrac{Q'}{C}$ , and transferring a further $\delta Q'$ requires work

$\delta W=V'\,\delta Q'=\dfrac{Q'}{C}\,\delta Q'$

Integrating from $0$ to $Q$ :

$W=\int_0^{Q}\dfrac{Q'}{C}\,dQ'=\dfrac{Q^{2}}{2C}$

Since the electrostatic force is conservative, this work is stored as potential energy of the charged capacitor, and the result does not depend on how the charge was built up.

$U=\dfrac{Q^{2}}{2C}=\dfrac{1}{2}CV^{2}=\dfrac{1}{2}QV$

Energy density

Write the energy of a parallel plate capacitor using $C=\dfrac{\varepsilon_0A}{d}$ and $Q=\sigma A$ :

$U=\dfrac{Q^{2}}{2C}=\dfrac{(\sigma A)^{2}}{2}\cdot\dfrac{d}{\varepsilon_0A}=\dfrac{\sigma^{2}}{2\varepsilon_0}(Ad)$

and since $E=\dfrac{\sigma}{\varepsilon_0}$ ,

$U=\dfrac{1}{2}\varepsilon_0E^{2}(Ad)$

The factor $Ad$ is the volume of the gap, where the field lives, so the energy per unit volume is

$u=\dfrac{1}{2}\varepsilon_0E^{2}$

Unit $\text{J m}^{-3}$ . Although derived for parallel plates, the result is general: it holds for the field of any charge configuration.
With a dielectric of constant $K$ filling the region, $u=\dfrac12 K\varepsilon_0E^{2}$ .

Choosing which form to use

Battery still connected: $V$ is fixed, so use $U=\dfrac12CV^{2}$ . If $C\to KC$ then $U\to KU$ .
Battery disconnected: $Q$ is fixed, so use $U=\dfrac{Q^{2}}{2C}$ . If $C\to KC$ then $U\to\dfrac{U}{K}$ .
Series: same $Q$ , so $\dfrac{U_1}{U_2}=\dfrac{C_2}{C_1}$ . Parallel: same $V$ , so $\dfrac{U_1}{U_2}=\dfrac{C_1}{C_2}$ .

Energy of a combination equals the sum of the parts

For three capacitors in series across $V$ , the charge $Q$ is common and $\dfrac{1}{C_{\text{eff}}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}$ , so

$U=\dfrac{Q^{2}}{2C_{\text{eff}}}=\dfrac{Q^{2}}{2C_1}+\dfrac{Q^{2}}{2C_2}+\dfrac{Q^{2}}{2C_3}=U_1+U_2+U_3$

The same holds in parallel, using $V$ common and $\dfrac12CV^{2}$ .

Loss of energy when charge is shared

A charged capacitor connected across an identical uncharged one:

Charge is conserved: total $Q$ spread over $2C$ , so the common potential is $V'=\dfrac{Q}{2C}=\dfrac{V}{2}$ .
Initial energy $U_i=\dfrac{Q^{2}}{2C}$ ; final energy $U_f=\dfrac12(2C)\left(\dfrac{Q}{2C}\right)^{2}=\dfrac{Q^{2}}{4C}$ .

$\dfrac{U_f}{U_i}=\dfrac12$

so exactly half the energy disappears. It is not lost from the universe: a transient current flows between the capacitors while the potentials equalise, and the energy goes out as heat in the connecting wires and as electromagnetic radiation. For unequal capacitors the loss is $\dfrac{C_1C_2(V_1-V_2)^{2}}{2(C_1+C_2)}$ .

NCERT's numbers: a $900\ \text{pF}$ capacitor charged to $100\ \text{V}$ stores $\dfrac12CV^{2}=4.5\times10^{-6}\ \text{J}$ ; connected to an identical uncharged one, the system stores $2.25\times10^{-6}\ \text{J}$ .

Worked patterns CBSE reuses

Three capacitors in series between $90\ \text{V}$ and earth, $C_1=20$ , $C_2=30$ , $C_3=15\ \mu\text{F}$ : $\dfrac{1}{C_{eq}}=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{15}=\dfrac{9}{60}$ , so $C_{eq}=\dfrac{60}{9}\ \mu\text{F}$ and $Q=C_{eq}V=600\ \mu\text{C}$ . Then $V_2=\dfrac{Q}{C_2}=20\ \text{V}$ and $U_2=\dfrac12C_2V_2^{2}=6\times10^{-3}\ \text{J}$ .
Two $12\ \text{pF}$ capacitors on $50\ \text{V}$ : series gives $C_s=6\ \text{pF}$ , $U_s=7.5\ \text{nJ}$ , $Q_s=300\ \text{pC}$ ; parallel gives $C_p=24\ \text{pF}$ , $U_p=30\ \text{nJ}$ , $Q_p=1200\ \text{pC}$ .
$12\ \text{pF}$ alone on $50\ \text{V}$ : $U=\dfrac12CV^{2}=1.5\times10^{-8}\ \text{J}$ . With $6\ \text{pF}$ added in series across the same battery: $Q=\dfrac{C_1C_2}{C_1+C_2}V=200\ \text{pC}$ , so $V_1=16.67\ \text{V}$ across the $12\ \text{pF}$ and $V_2=33.33\ \text{V}$ across the $6\ \text{pF}$ .
Two capacitors, series energy $0.045\ \text{J}$ and parallel energy $0.25\ \text{J}$ at $100\ \text{V}$ : $C_1+C_2=5\times10^{-5}\ \text{F}$ and $C_1C_2=4.5\times10^{-10}$ , so $C_1-C_2=\sqrt{(C_1+C_2)^{2}-4C_1C_2}=2.64\times10^{-5}$ , giving $C_1=38.2\ \mu\text{F}$ and $C_2=11.8\ \mu\text{F}$ ; in parallel $Q_1=38.2\times10^{-4}\ \text{C}$ and $Q_2=11.8\times10^{-4}\ \text{C}$ .
Equal energies from a parallel and a series pair with $C_1:C_2=1:2$ : $C_p=3C_1$ , $C_s=\dfrac{2C_1}{3}$ , and $\dfrac12C_sV_s^{2}=\dfrac12C_pV_p^{2}$ gives $\dfrac{V_s}{V_p}=\dfrac{3}{\sqrt2}$ .
$X$ (air) in series with $Y$ ( $\varepsilon_r=4$ ), same $A$ and $d$ , equivalent $4\ \mu\text{F}$ across $15\ \text{V}$ : $C_y=4C_x$ and $\dfrac14=\dfrac{1}{C_x}+\dfrac{1}{4C_x}$ gives $C_x=5\ \mu\text{F}$ , $C_y=20\ \mu\text{F}$ ; $Q=60\ \mu\text{C}$ , $V_x=12\ \text{V}$ , $V_y=3\ \text{V}$ , and $\dfrac{U_x}{U_y}=\dfrac{C_y}{C_x}=4$ .
Two identical capacitors on a battery with a switch: with the switch closed both are at $V$ , $U_i=CV^{2}$ . Open the switch (isolating $B$ at fixed charge, leaving $A$ at fixed $V$ ) and insert $K$ in both: $U_A'=\dfrac12KCV^{2}$ , $U_B'=\dfrac{(CV)^{2}}{2KC}=\dfrac{CV^{2}}{2K}$ , so $\dfrac{U_i}{U_f}=\dfrac{2K}{K^{2}+1}$ .
A variable capacitor on a $200\ \text{V}$ battery changed from $2\ \mu\text{F}$ to $X$ , with the energy falling by $2\times10^{-2}\ \text{J}$ : $2\times10^{-2}=\dfrac12(200)^{2}(2-X)\times10^{-6}$ gives $X=1\ \mu\text{F}$ .

Diagram to describe in words

Two conductor plates, with an arrow showing a small charge $\delta Q'$ being carried from the lower plate to the upper against the existing potential difference $V'$ . Beside it, a graph of $V'$ against $Q'$ — a straight line through the origin of slope $1/C$ — with the triangular area beneath it shaded and labelled $\dfrac{Q^{2}}{2C}$ , since the stored energy is that area.

Standard results

Quantity	Result
Energy stored	$\dfrac12CV^{2}=\dfrac12QV=\dfrac{Q^{2}}{2C}$
Energy density	$u=\dfrac12\varepsilon_0E^{2}$
With dielectric $K$	$u=\dfrac12K\varepsilon_0E^{2}$
Energy of a combination	Sum of the individual energies
Sharing with an identical uncharged capacitor	$\dfrac{U_f}{U_i}=\dfrac12$
Loss on sharing, general	$\dfrac{C_1C_2(V_1-V_2)^{2}}{2(C_1+C_2)}$
Battery connected, insert $K$	$U\to KU$
Battery disconnected, insert $K$	$U\to\dfrac{U}{K}$

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
Calculate the potential difference and the energy stored in the capacitor $C_2$ in the circuit shown in the figure. Given potential at $A$ is 90 V, $C_1=20\,\mu\text{F}$ , $C_2=30\,\mu\text{F}$ and $C_3=15\,\mu\text{F}$ .
2 marksCBSE Class 12 Physics — SA · AI 2015, An · Section B, 2 marksAnswer 1 ↓
2
A parallel plate capacitor of capacitance $C$ is charged to a potential $V$ . It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.
2 marksCBSE Class 12 Physics — SA · AI 2014 · Section B, 2 marksAnswer 2 ↓
3
In the figure given below, find the (a) equivalent capacitance of the network between points $A$ and $B$ , given $C_1=C_5=4\,\mu\text{F}$ , $C_2=C_3=C_4=2\,\mu\text{F}$ ; (b) maximum charge supplied by the battery; and (c) total energy stored in the network.
3 marksCBSE Class 12 Physics — SA · 2020, An · Section C, 3 marksAnswer 3 ↓
4
(i) Find the equivalent capacitance between $A$ and $B$ in the combination given below. Each capacitor is of $2\,\mu\text{F}$ capacitance. (ii) If a dc source of 7 V is connected across $AB$ , how much charge is drawn from the source and what is the energy stored in the network?
3 marksCBSE Class 12 Physics — SA · Delhi 2017 · Section C, 3 marksAnswer 4 ↓
5
A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor? If another capacitor of 6 pF is connected in series with it with the same battery connected across the combination, find the charge stored and potential difference across each capacitor.
3 marksCBSE Class 12 Physics — SA · Delhi 2017, Ap · Section C, 3 marksAnswer 5 ↓
6
Two identical capacitors of 12 pF each are connected in series across a battery of 50 V. How much electrostatic energy is stored in the combination? If these were connected in parallel across the same battery, how much energy will be stored in the combination now? Also find the charge drawn from the battery in each case.
5 marksCBSE Class 12 Physics — Long Answer · Delhi 2017 · Section E, 5 marksAnswer 6 ↓
7
Two identical parallel plate capacitors $A$ and $B$ are connected to a battery of $V$ volt with the switch $S$ closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant $K$ . Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.
5 marksCBSE Class 12 Physics — Long Answer · AI 2017, An · Section E, 5 marksAnswer 7 ↓
8
Two parallel plate capacitors $X$ and $Y$ have the same area of plates and same separation between them. $X$ has air between the plates while $Y$ contains a dielectric of $\varepsilon_r=4$ . (i) Calculate the capacitance of each capacitor if the equivalent capacitance of the combination is $4\,\mu\text{F}$ . (ii) Calculate the potential difference between the plates of $X$ and $Y$ . (iii) Estimate the ratio of electrostatic energy stored in $X$ and $Y$ . ( $X$ and $Y$ are connected in series across a 15 V battery.)
5 marksCBSE Class 12 Physics — Long Answer · Delhi 2016 · Section E, 5 marksAnswer 8 ↓
9
In the following arrangement of capacitors, the energy stored in the $6\,\mu\text{F}$ capacitor is $E$ . Find the value of (i) energy stored in the $12\,\mu\text{F}$ capacitor, (ii) energy stored in the $3\,\mu\text{F}$ capacitor, (iii) total energy drawn from the battery.
5 marksCBSE Class 12 Physics — Long Answer · Foreign 2016, Ap · Section E, 5 marksAnswer 9 ↓
10
Two capacitors of unknown capacitances $C_1$ and $C_2$ are connected first in series and then in parallel across a battery of 100 V. If the energy stored in the two combinations is 0.045 J and 0.25 J respectively, determine the value of $C_1$ and $C_2$ . Also calculate the charge on each capacitor in parallel combination.
5 marksCBSE Class 12 Physics — Long Answer · Delhi 2015 · Section E, 5 marksAnswer 10 ↓
11
(i)(A) Why does the electric field inside a dielectric slab decrease when kept in an external electric field? (B) Derive an expression for the capacitance of a parallel plate capacitor filled with a medium of dielectric constant $K$ . (ii) A charge $q=2\,\mu\text{C}$ is placed at the centre of a sphere of radius 20 cm. What is the amount of work done in moving $4\,\mu\text{C}$ from one point on its surface to another point? (iii) Write a relation for polarisation $\vec P$ of a dielectric material in the presence of an external electric field.
5 marksCBSE Class 12 Physics — Long Answer · AI 2021C · Section E, 5 marksAnswer 11 ↓
12
(i) Obtain an expression for the potential energy of an electric dipole placed in a uniform electric field. (ii) Three capacitors of capacitance $C_1$ , $C_2$ and $C_3$ are connected in series to a source of $V$ volt. Show that the total energy stored in the combination of capacitors is equal to the sum of the energy stored in the individual capacitors. (iii) A capacitor of capacitance $C$ is connected across a battery. After charging, the battery is disconnected and the separation between the plates is doubled. How will (a) the capacitance and (b) the electric field between the plates be affected? Justify your answer.
5 marksCBSE Class 12 Physics — Long Answer · AI 2021C · Section E, 5 marksAnswer 12 ↓
13
(a) Obtain the expressions for the resultant capacitance when three capacitors $C_1$ , $C_2$ and $C_3$ are connected (i) in parallel and then (ii) in series. (b) In the circuit shown in the figure, the charge on the capacitor of $4\,\mu\text{F}$ is $16\,\mu\text{C}$ . Calculate the energy stored in the capacitor of $12\,\mu\text{F}$ capacitance.
5 marksCBSE Class 12 Physics — Long Answer · AI 2019C · Section E, 5 marksAnswer 13 ↓
14
(a) When a parallel plate capacitor is connected across a dc battery, explain briefly how the capacitor gets charged. (b) A parallel plate capacitor of capacitance $C$ is charged to $V$ volt by a battery. After some time the battery is disconnected and the distance between the plates is doubled. Now a slab of dielectric constant $1<K<2$ is introduced to fill the space between the plates. How will the following be affected? (i) The electric field between the plates of the capacitor. (ii) The energy stored in the capacitor. Justify your answer in each case.
5 marksCBSE Class 12 Physics — Long Answer · 2/5, AI 2019, Ev · Section E, 5 marksAnswer 14 ↓
15
A parallel plate capacitor is charged by a battery to a potential difference $V$ . It is disconnected from the battery and then connected to another uncharged capacitor of the same capacitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single capacitor.
3 marksCBSE Class 12 Physics — SA · 2/5, Delhi 2019 · Section C, 3 marksAnswer 15 ↓
16
A parallel plate capacitor of capacitance $C$ is charged to $V$ volt by a battery. After some time the battery is disconnected and the distance between the plates is doubled. Now a slab of dielectric constant $1<K<2$ is introduced to fill the space between the plates. How will the following be affected? (i) The electric field between the plates of the capacitor. (ii) The energy stored in the capacitor. Justify your answer in each case.
3 marksCBSE Class 12 Physics — SA · 2/5, AI 2019, Ev · Section C, 3 marksAnswer 16 ↓
17
Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors $C_1$ and $C_2$ with their capacitances in the ratio $1:2$ so that the energy stored in the two cases becomes the same.
5 marksCBSE Class 12 Physics — Long Answer · 3/5, AI 2016, Ev · Section E, 5 marksAnswer 17 ↓
18
A fully charged parallel plate capacitor is connected across an uncharged identical capacitor. Show that the energy stored in the combination is less than that stored initially in the single capacitor.
5 marksCBSE Class 12 Physics — Long Answer · AI 2015 · Section E, 5 marksAnswer 18 ↓

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

19
A variable capacitor is connected to a 200 V battery. If its capacitance is changed from $2\,\mu\text{F}$ to $X\,\mu\text{F}$ , the decrease in energy of the capacitor is $2\times10^{-2}\,\text{J}$ . The value of $X$ is
- (a)
  $1\,\mu\text{F}$
- (b)
  $2\,\mu\text{F}$
- (c)
  $3\,\mu\text{F}$
- (d)
  $4\,\mu\text{F}$
1 markCBSE Class 12 Physics — Multiple Choice · Term I 2021-22 · Section A, 1 markAnswer 19 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.↑ back to the question
The three capacitors are in series between $A$ (90 V) and $B$ (grounded, 0 V). Equivalent capacitance: $\dfrac{1}{C_{eq}}=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{15}=\dfrac{3+2+4}{60}=\dfrac{9}{60} \Rightarrow C_{eq}=\dfrac{60}{9}\,\mu\text{F}$ . Charge on equivalent (same on each, series): $Q=C_{eq}V=\dfrac{60}{9}\times10^{-6}\times90=600\,\mu\text{C}$ . Potential drop across $C_2$ : $V_2=\dfrac{Q}{C_2}=\dfrac{600\times10^{-6}}{30\times10^{-6}}=20\,\text{V}$ . Energy stored, $U=\dfrac12 C_2V_2^2=\dfrac12\times30\times10^{-6}\times(20)^2=6\times10^{-3}\,\text{J}$ .
2.↑ back to the question
Total charge $Q=CV$ is conserved. Combined (parallel) capacitance $=C+C=2C$ . Initial energy (single capacitor) $=\dfrac{1}{2}\dfrac{Q^2}{C}$ . Final energy (combined) $=\dfrac{1}{2}\dfrac{Q^2}{2C}$ . Ratio $=\dfrac{\text{Final energy}}{\text{Initial energy}}=\dfrac{1}{2}$ .
3.↑ back to the question
(a) $C_1,C_2$ in series ( $AB$ upper-left branch) $=\dfrac{4\times2}{4+2}=\dfrac43\,\mu\text{F}$ ; $C_4,C_5$ in series (upper-right branch) $=\dfrac{2\times4}{2+4}=\dfrac43\,\mu\text{F}$ . These two combine in parallel with the direct wire through $C_3$ 's node... Using the standard reduction for this bridge-type network with the given values: $C_{eq}\approx2\,\mu\text{F}$ (equivalent capacitance of the network between $A$ and $B$ ). (b) Maximum charge from the 5 V battery: $Q_{max}=C_{eq}V=2\times10^{-6}\times5=10\,\mu\text{C}$ . (c) Total energy stored: $E=\dfrac12 C_{eq}V^2=\dfrac12\times2\times10^{-6}\times(5)^2=25\,\mu\text{J}$ .
4.↑ back to the question
(i) $C_2$ , $C_3$ and $C_4$ (the middle bridging branches $P$ - $R$ - $S$ - $T$ ) are in parallel: $C_{\text{parallel}}=C_2+C_3+C_4=2+2+2=6\,\mu\text{F}$ . This parallel combination is in series with $C_1$ and $C_5$ : $\dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_{\text{parallel}}}+\dfrac{1}{C_5}=\dfrac12+\dfrac16+\dfrac12=\dfrac{3+1+3}{6}=\dfrac76 \Rightarrow C_{eq}=\dfrac67=0.86\,\mu\text{F}$ . (ii) $Q=C_{eq}V=0.86\times7=6\,\mu\text{C}$ (charge drawn from the source). Energy, $E=\dfrac12QV=\dfrac12\times6\times7=21\,\mu\text{J}$ .
5.↑ back to the question
Energy stored, $U=\dfrac12CV^2=\dfrac12\times12\times10^{-12}\times(50)^2=1.5\times10^{-8}\,\text{J}$ . With 6 pF in series with 12 pF: charge on each (series, same charge), $Q=\dfrac{C_1C_2}{C_1+C_2}V=\dfrac{12\times6\times10^{-24}}{18\times10^{-12}}\times50=200\,\text{pC}$ . Potential difference across 12 pF: $V_1=\dfrac{Q}{C_1}=\dfrac{200\times10^{-12}}{12\times10^{-12}}=16.67\,\text{V}$ . Across 6 pF: $V_2=\dfrac{Q}{C_2}=\dfrac{200\times10^{-12}}{6\times10^{-12}}=33.33\,\text{V}$ .
6.↑ back to the question
Series: $C_s=\dfrac{12\times12}{12+12}=6\,\text{pF}$ ; $U_s=\dfrac12C_sV^2=\dfrac12\times6\times10^{-12}\times(50)^2=7.5\,\text{nJ}$ . Parallel: $C_p=12+12=24\,\text{pF}$ ; $U_p=\dfrac12C_pV^2=\dfrac12\times24\times10^{-12}\times(50)^2=30\,\text{nJ}$ . Charge from battery (series): $Q_s=C_sV=6\times10^{-12}\times50=300\,\text{pC}$ . Charge from battery (parallel): $Q_p=C_pV=24\times10^{-12}\times50=1200\,\text{pC}$ .
7.↑ back to the question
Initially (switch closed), both $A$ and $B$ (capacitance $C$ each) are in parallel across the battery, so total initial energy $U_i=2\times\dfrac12CV^2=CV^2$ . When $S$ is opened, $B$ is isolated (charge $Q=CV$ on it remains constant); $A$ stays connected to the battery (potential $V$ constant). After the dielectric ( $K$ ) is inserted: capacitance of each becomes $KC$ . For $A$ (constant $V$ ): $U_A'=\dfrac12KCV^2$ . For $B$ (constant $Q$ ): $U_B'=\dfrac{Q^2}{2KC}=\dfrac{(CV)^2}{2KC}=\dfrac{CV^2}{2K}$ . Total final energy, $U_f=\dfrac12KCV^2+\dfrac{CV^2}{2K}=\dfrac12CV^2\left(K+\dfrac1K\right)$ . Ratio, $\dfrac{U_i}{U_f}=\dfrac{CV^2}{\tfrac12CV^2(K+1/K)}=\dfrac{2K}{K^2+1}$ .
8.↑ back to the question
(i) Let $C_x=\dfrac{\varepsilon_0A}{d}$ ; $C_y=\varepsilon_rC_x=4C_x$ . In series: $\dfrac{1}{4}=\dfrac{1}{C_x}+\dfrac{1}{4C_x}=\dfrac{5}{4C_x}\Rightarrow C_x=5\,\mu\text{F}$ and $C_y=4C_x=20\,\mu\text{F}$ . (ii) Charge (same, series): $Q=C_{eq}V=4\times10^{-6}\times15=60\,\mu\text{C}$ . $V_x=\dfrac{Q}{C_x}=\dfrac{60}{5}=12\,\text{V}$ ; $V_y=V-V_x=15-12=3\,\text{V}$ . (iii) $\dfrac{U_x}{U_y}=\dfrac{Q^2/2C_x}{Q^2/2C_y}=\dfrac{C_y}{C_x}=\dfrac{4C_x}{C_x}=4$ .
9.↑ back to the question
(i) $6\,\mu\text{F}$ and $12\,\mu\text{F}$ are in parallel with each other, so they share the same potential $V$ : $\dfrac12\times6\times V^2=E\Rightarrow V^2=\dfrac{E}{3}\times10^6$ (schematically). $E_{12}=\dfrac12\times12\times V^2=2\times\left(\dfrac12\times6\times V^2\right)=2E$ . (ii) The $(6\parallel12)$ combination is in series with $3\,\mu\text{F}$ , so the same charge flows through $3\,\mu\text{F}$ as the combined charge of the parallel pair: working through the series-charge relation gives $E_3=18E$ . (iii) Total energy drawn from battery, $E_{\text{total}}=E+E_{12}+E_3=E+2E+18E=21E$ .
10.↑ back to the question
Parallel: $U_p=\dfrac12(C_1+C_2)V^2=0.25\Rightarrow C_1+C_2=\dfrac{2\times0.25}{(100)^2}=5\times10^{-5}\,\text{F}$ ...(i) Series: $U_s=\dfrac12\dfrac{C_1C_2}{C_1+C_2}V^2=0.045\Rightarrow C_1C_2=\dfrac{2\times0.045\times(C_1+C_2)}{(100)^2}=4.5\times10^{-10}$ ...(ii) $(C_1-C_2)=\sqrt{(C_1+C_2)^2-4C_1C_2}=\sqrt{(5\times10^{-5})^2-4\times4.5\times10^{-10}}=2.64\times10^{-5}$ . Solving with (i): $C_1=38.2\,\mu\text{F}$ , $C_2=11.8\,\mu\text{F}$ . In parallel, charge on each: $Q_1=C_1V=38.2\times10^{-6}\times100=38.2\times10^{-4}\,\text{C}$ ; $Q_2=C_2V=11.8\times10^{-6}\times100=11.8\times10^{-4}\,\text{C}$ .
11.↑ back to the question
Answer: (i) (a); (i) (b) (i)(A) A dielectric gets polarised in an external field; the field produced by the induced (bound) surface charges opposes the external field, so the net field inside the dielectric is reduced. (i)(B) With dielectric $K$ filling the gap, $E=\dfrac{E_0}{K}=\dfrac{\sigma}{K\varepsilon_0}$ ; potential difference $V=Ed=\dfrac{\sigma d}{K\varepsilon_0}$ ; $C=\dfrac{Q}{V}=\dfrac{Q K\varepsilon_0}{\sigma d}=\dfrac{K\varepsilon_0A}{d}$ (using $Q=\sigma A$ ). (ii) Since the sphere's surface is equipotential, the work done in moving a charge from one point on the surface to another is $W=q\Delta V=0$ . (iii) $\vec P=\chi_e\vec E$ , where $\chi_e$ is the electric susceptibility of the dielectric.
12.↑ back to the question
(i) $U=-pE\cos\theta=-\vec p\cdot\vec E$ (as derived earlier). (ii) In series, $\dfrac{1}{C_{eff}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}$ , and the charge $Q$ on each capacitor is the same. Total energy, $U=\dfrac{Q^2}{2C_{eff}}=\dfrac{Q^2}{2C_1}+\dfrac{Q^2}{2C_2}+\dfrac{Q^2}{2C_3}=U_1+U_2+U_3$ . (iii) (a) With separation doubled ( $d\to2d$ ), $C'=\dfrac{\varepsilon_0A}{2d}=\dfrac{C}{2}$ — capacitance halves. (b) Since charge $Q$ remains constant (battery disconnected) and field $E=\dfrac{\sigma}{\varepsilon_0}=\dfrac{Q}{\varepsilon_0A}$ does not depend on separation, the field between the plates remains unchanged.
13.↑ back to the question
(a)(i) Parallel: $q=q_1+q_2+q_3=C_1V+C_2V+C_3V \Rightarrow C_p=C_1+C_2+C_3$ . (ii) Series: $V=V_1+V_2+V_3=\dfrac{q}{C_1}+\dfrac{q}{C_2}+\dfrac{q}{C_3} \Rightarrow \dfrac{1}{C_s}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}$ . (b) $20\,\mu\text{F}$ and $4\,\mu\text{F}$ are in parallel: $C_p=24\,\mu\text{F}$ , with charge on $4\,\mu\text{F}$ given as $16\,\mu\text{C}$ , so $V_{4\mu F}=\dfrac{16}{4}=4\,\text{V}=V_{20\mu F\parallel}$ . Charge on the $(20\parallel4)$ combination: $q'=24\times4=96\,\mu\text{C}$ — this equals the charge on $12\,\mu\text{F}$ (in series with it, from the 12 V source): $V_{12\mu F}=\dfrac{96}{12}=8\,\text{V}$ (consistent with $12=8+4$ ). Energy in $12\,\mu\text{F}$ : $U=\dfrac12\times12\times10^{-6}\times(8)^2=384\,\mu\text{J}$ .
14.↑ back to the question
(a) Same explanation as Q40: current flows momentarily, charges accumulate on the plates until the potential difference equals the battery voltage. (b)(i) Before dielectric: field with doubled separation, charge $Q$ fixed, $E_0=\dfrac{\sigma}{\varepsilon_0}$ (independent of $d$ , unaffected by doubling). With dielectric $K$ inserted: $E=\dfrac{E_0}{K}$ — field decreases by factor $K$ from the value before the dielectric was added (though independent of $d$ itself). (ii) Energy $U=\dfrac{Q^2}{2C}$ ; capacitance with doubled separation and dielectric: $C'=\dfrac{K\varepsilon_0A}{2d}$ . So $U'=\dfrac{Q^2}{2C'}=\dfrac{Q^2\times2d}{2K\varepsilon_0A}=\dfrac{1}{K}\times\dfrac{Q^2d}{\varepsilon_0A}$ — since $Q$ is fixed and $C'<C$ (as $1<K<2$ still gives less capacitance than the original due to doubled $d$ dominating), the energy stored increases compared to a capacitor without the dielectric at the doubled separation, but decreases compared to before separation was doubled overall, the field decreasing and capacitance decreasing means energy $=Q^2/2C$ increases as $C$ decreases.
15.↑ back to the question
Same result as Q50: total charge conserved, combined capacitance doubles, so $\dfrac{U_f}{U_i}=\dfrac12$ .
16.↑ back to the question
(i) With separation doubled: $E'=\dfrac{V'}{d'}=\dfrac{V}{d}\times\dfrac{1}{2}=\dfrac12E$ (before dielectric). With dielectric $K$ also introduced: $E''=\dfrac{E'}{K}=\dfrac{E}{2K}$ — the field reduces to half (from doubling $d$ with $Q$ fixed... since $E=\sigma/\varepsilon_0$ independent of $d$ for fixed charge, more precisely) — the net result (per the book) is that the field reduces to $\dfrac12\left(\dfrac{E}{K}\right)$ of its original value. (ii) New capacitance, $C'=\dfrac{K\varepsilon_0A}{2d}=\dfrac{K}{2}C$ . Since $Q$ (fixed) is unchanged: $U'=\dfrac{Q^2}{2C'}=\dfrac{Q^2}{2(K/2)C}=\dfrac{2}{K}\left(\dfrac{Q^2}{2C}\right)=\dfrac{2U}{K}$ . Since $1<K<2$ , $\dfrac2K>1$ , so the energy stored increases.
17.↑ back to the question
Given $C_1:C_2=1:2$ , so $C_2=2C_1$ . Parallel: $C_p=C_1+C_2=3C_1$ . Series: $\dfrac{1}{C_s}=\dfrac{1}{C_1}+\dfrac{1}{2C_1}=\dfrac{3}{2C_1}\Rightarrow C_s=\dfrac{2C_1}{3}$ . Equal energies: $\dfrac12C_sV_s^2=\dfrac12C_pV_p^2 \Rightarrow \dfrac{2}{3}C_1V_s^2=3C_1V_p^2 \Rightarrow \dfrac{V_s^2}{V_p^2}=\dfrac{9}{2} \Rightarrow \dfrac{V_s}{V_p}=\dfrac{3}{\sqrt2}.$
18.↑ back to the question
Let the fully charged capacitor $C$ have charge $Q$ , so initial energy $U=\dfrac{Q^2}{2C}$ . When connected to an identical uncharged capacitor $C$ , charge redistributes equally by symmetry; total capacitance in parallel $=2C$ , and total charge $Q$ is conserved. Common potential $V=\dfrac{Q}{2C}$ . Final energy, $U'=\dfrac12(2C)V^2=\dfrac12(2C)\left(\dfrac{Q}{2C}\right)^2=\dfrac{Q^2}{4C}$ . Since $U'=\dfrac{Q^2}{4C}<\dfrac{Q^2}{2C}=U$ , the combined energy is less than the initial energy (the difference is dissipated as heat and electromagnetic radiation during the charge redistribution).
19.Answer: (A)↑ back to the question
Correct option: (a). $\Delta U=\dfrac12 C_1V^2-\dfrac12 C_2V^2$ : $2\times10^{-2}=\dfrac12\times200\times200\times(2-X)\times10^{-6}$ . $\Rightarrow (2-X)=1 \Rightarrow X=1\,\mu\text{F}$ .