Chapter 2 · Electrostatic Potential and Capacitance

2.4 Combinations of Capacitors and Stored Energy

Capacitors in series share the same charge and $\dfrac1C=\sum\dfrac1{C_i}$; in parallel they share the same voltage and $C=\sum C_i$. A charged capacitor stores energy $U=\dfrac{Q^2}{2C}=\tfrac12CV^2$, held in its field at $\tfrac12\varepsilon_0E^2$ per unit volume.

A heart defibrillator must deliver a carefully measured jolt — typically $150$ to $360$ joules — in a few milliseconds. No battery can give up its energy that fast, but a capacitor can: the battery slowly charges a capacitor over several seconds, and the capacitor then dumps its energy through the patient's chest almost at once. Electric buses in Shanghai do something similar with "supercapacitors", recharging in seconds at each stop, and every power supply in your home smooths its output with capacitors.

Real circuits rarely use one capacitor alone. This section shows how to replace any combination of capacitors in series and in parallel by a single equivalent capacitor, and how charge and voltage divide between them. Then we calculate the energy stored in a charged capacitor, find that it can be thought of as stored in the electric field itself, and follow what happens to charge and energy when capacitors are connected to each other.

Capacitors in Series

Connect capacitors $C_1$, $C_2$, $C_3$ one after another across a battery of voltage $V$. The battery moves charge $Q$ from the right-hand plate of $C_3$ to the left-hand plate of $C_1$. The inner plates were joined only to each other and started neutral, so they stay neutral as a pair: $+Q$ on one induces $-Q$ on the facing plate, which leaves $+Q$ on the next. Every capacitor in series carries the same charge $Q$, while the voltages add: $V=V_1+V_2+V_3=\dfrac{Q}{C_1}+\dfrac{Q}{C_2}+\dfrac{Q}{C_3}$.

📐 Result 2.13 — Capacitors in series

Capacitors connected in series (end to end, with no branch at any junction) carry equal charges $Q$ and behave as a single capacitor of capacitance $C$ given by $$\frac1C=\frac1{C_1}+\frac1{C_2}+\cdots+\frac1{C_n}.$$ The equivalent capacitance is smaller than the smallest one in the chain. The voltage divides in inverse proportion to capacitance, $V_i=Q/C_i$: the smallest capacitor takes the largest share. For two capacitors, $C=\dfrac{C_1C_2}{C_1+C_2}$.

✏️ Example 1 — Three capacitors in series
Capacitors of $2.0\ \mu\text{F}$, $3.0\ \mu\text{F}$ and $6.0\ \mu\text{F}$ are connected in series across a $12$ V battery. Find the equivalent capacitance, the charge on each capacitor and the voltage across each.
Equivalent $\dfrac1C=\dfrac12+\dfrac13+\dfrac16=\dfrac{3+2+1}{6}=1\ \mu\text{F}^{-1}$, so $C=1.0\ \mu\text{F}$ — less than the smallest, $2.0\ \mu\text{F}$.
Charge $Q=CV=(1.0\ \mu\text{F})(12\ \text{V})=12\ \mu\text{C}$ on each capacitor.
Voltages $V_1=\dfrac{12}{2.0}=6.0$ V, $V_2=\dfrac{12}{3.0}=4.0$ V, $V_3=\dfrac{12}{6.0}=2.0$ V.
Check $6.0+4.0+2.0=12$ V ✓. Choose "all in series" in Interactive 1 with these values to see the same numbers.

Capacitors in Parallel

Now connect the capacitors side by side, each one directly across the battery. Each then has the full battery voltage $V$ across it and takes its own charge $Q_i=C_iV$; the battery supplies the total, $Q=Q_1+Q_2+Q_3=(C_1+C_2+C_3)V$.

📐 Result 2.14 — Capacitors in parallel

Capacitors connected in parallel (all between the same two points) have the same voltage $V$ and behave as a single capacitor of capacitance $$C=C_1+C_2+\cdots+C_n .$$ The charge divides in direct proportion to capacitance, $Q_i=C_iV$: the largest capacitor takes the most charge. In effect the plate areas add.

🧭 Problem-solving strategy — reducing a capacitor network
  1. Find a group that is purely in series (same charge — nothing branches off between them) or purely in parallel (same two end points). Replace it by its equivalent and redraw.
  2. Repeat until a single capacitor remains. Then $Q_\text{total}=C_\text{eq}V$.
  3. Work backwards through your drawings: a series group passes the same charge to each member; a parallel group puts the same voltage across each member.
  4. Check: voltages round any loop add up to the battery voltage, and charges at each junction balance.
✏️ Example 2 — A series–parallel network
A $4.0\ \mu\text{F}$ capacitor $C_1$ is connected in series with a parallel pair $C_2=2.0\ \mu\text{F}$ and $C_3=6.0\ \mu\text{F}$, and the network is connected to a $24$ V battery. Find the equivalent capacitance and the charge on and voltage across each capacitor.
Reduce Parallel pair: $C_{23}=2.0+6.0=8.0\ \mu\text{F}$. In series with $C_1$: $C=\dfrac{(4.0)(8.0)}{4.0+8.0}=\dfrac{32}{12}\approx2.67\ \mu\text{F}$.
Total charge $Q=CV=\dfrac{32}{12}\times24=64\ \mu\text{C}$. This is the charge on $C_1$ and on the pair as a whole.
Voltages $V_1=\dfrac{64}{4.0}=16$ V; across the pair $V_{23}=\dfrac{64}{8.0}=8.0$ V (and $16+8=24$ ✓).
Split $Q_2=C_2V_{23}=16\ \mu\text{C}$, $Q_3=C_3V_{23}=48\ \mu\text{C}$; $16+48=64\ \mu\text{C}$ ✓. Interactive 1 opens on this network.
✏️ Example 3 — Designing with a voltage rating
You have a box of identical capacitors, each $2.0\ \mu\text{F}$ and rated for at most $50$ V. Design a combination that gives $2.0\ \mu\text{F}$ and can safely be used at $100$ V, using as few capacitors as possible.
Voltage At $100$ V each capacitor may take only $50$ V, so we need at least $2$ in series in every path. Two identical capacitors in series share the voltage equally: $50$ V each ✓, but the pair has only $\dfrac{2.0}{2}=1.0\ \mu\text{F}$.
Capacitance Put two such pairs in parallel: $1.0+1.0=2.0\ \mu\text{F}$ ✓, still with $50$ V across each capacitor.
Answer Four capacitors: two series pairs connected in parallel. In general $m$ in series and $n$ such rows in parallel give $C=\dfrac{n}{m}C_0$ with a rating of $m$ times the single rating.
Interactive 1 — Capacitor Network Builder

Pick a network and set the three capacitances and the battery voltage. The circuit replays its charging from zero (press Play / Pause): the $\pm$ signs grow with the charge, and each capacitor's label shows its charge and voltage at that moment; the readouts give the fully charged values. Notice which capacitors share the same $Q$ (series) and which share the same $V$ (parallel).

4.02.06.024
equivalent capacitance—
charge from battery—
$Q_1$, $Q_2$, $Q_3$—
$V_1$, $V_2$, $V_3$—
energy stored—

Energy Stored in a Capacitor

Charging a capacitor takes work, because each new bit of charge has to be pushed onto a plate that already repels it. Suppose that at some stage the plates carry $\pm q$, so the potential difference is $q/C$. Moving a further small charge $dq$ from the negative plate to the positive one takes work $dW=\dfrac qC\,dq$. Adding up from $q=0$ to $q=Q$ (uses integration): $$W=\int_0^Q\frac qC\,dq=\frac{Q^2}{2C}.$$ Geometrically, this is the area of the triangle under the straight-line graph of $V=q/C$ against $q$: base $Q$, height $V$, area $\tfrac12QV$. The average voltage during charging is only half the final voltage, which is why the factor $\tfrac12$ appears.

📐 Result 2.15 — Energy stored in a capacitor

A capacitor of capacitance $C$ with charge $Q$ and potential difference $V=Q/C$ stores electrostatic potential energy $$U=\frac{Q^2}{2C}=\frac12CV^2=\frac12QV .$$ Use $\dfrac{Q^2}{2C}$ when $Q$ is fixed (isolated capacitor) and $\tfrac12CV^2$ when $V$ is fixed (connected to a battery). For capacitors in series or parallel, the total energy is the sum of the separate energies, and equals $\tfrac12C_\text{eq}V^2$.

Where is the energy? For a parallel-plate capacitor $C=\varepsilon_0A/d$ and $V=Ed$, so $$U=\frac12\,\frac{\varepsilon_0A}{d}\,(Ed)^2=\left(\frac12\varepsilon_0E^2\right)(Ad).$$ $Ad$ is the volume between the plates, where the field is. It is natural to say that the energy is stored in the field, with a fixed amount per unit volume. This turns out to be true for every electric field, not just this one, and the idea becomes essential in Chapter 8, where electromagnetic waves carry energy through empty space.

📘 Definition 2.6 — Energy density of an electric field
The energy density $u$ of an electric field is the electrostatic energy stored per unit volume of space where the field is: $$u=\frac12\varepsilon_0E^2\quad\text{(vacuum)},\qquad u=\frac12K\varepsilon_0E^2\quad\text{(in a dielectric)} .$$ Its unit is J/m³.
✏️ Example 4 — The energy in a defibrillator
A defibrillator is to deliver $200$ J from a capacitor charged to $2.0$ kV. (a) What capacitance is needed? (b) What charge flows through the patient if the capacitor discharges completely? (c) If the discharge takes $5.0$ ms, what is the average power?
(a) Capacitance $U=\tfrac12CV^2\Rightarrow C=\dfrac{2U}{V^2}=\dfrac{2(200)}{(2.0\times10^3)^2}=1.0\times10^{-4}$ F $=100\ \mu\text{F}$.
(b) Charge $Q=CV=(1.0\times10^{-4})(2.0\times10^3)=0.20$ C. (Check: $\tfrac12QV=\tfrac12(0.20)(2000)=200$ J ✓.)
(c) Power $P=\dfrac{U}{t}=\dfrac{200}{5.0\times10^{-3}}=4.0\times10^4$ W $=40$ kW — far more than the charging circuit's battery could supply directly. Charging slowly and discharging quickly is the whole point.
✏️ Example 5 — Energy density between the plates
The air capacitor of Section 2.3, Example 5 ($A=1.0\times10^{-2}\ \text{m}^2$, $d=2.0$ mm, $C=44.3$ pF) is charged to $100$ V. Find (a) the energy stored, using $\tfrac12CV^2$; (b) the energy density of the field, and check that it accounts for all of the energy.
(a) $U=\tfrac12(4.43\times10^{-11})(100)^2\approx2.2\times10^{-7}$ J.
(b) $E=\dfrac Vd=5.0\times10^4$ V/m, so $u=\tfrac12(8.85\times10^{-12})(5.0\times10^4)^2\approx1.1\times10^{-2}\ \text{J/m}^3$.
Check Volume $Ad=2.0\times10^{-5}\ \text{m}^3$; $u\times Ad=(1.106\times10^{-2})(2.0\times10^{-5})\approx2.2\times10^{-7}$ J ✓ — all of the energy is in the uniform field between the plates.

Sharing Charge between Capacitors

Connect a charged capacitor $C_1$ (at voltage $V_1$) to another capacitor $C_2$ (at $V_2$, perhaps uncharged), positive plate to positive plate. Charge flows through the wires until both are at the same common potential $V$. No charge can leave the isolated pair of connected plates, so the total charge is conserved: $$C_1V_1+C_2V_2=(C_1+C_2)V\quad\Rightarrow\quad V=\frac{C_1V_1+C_2V_2}{C_1+C_2}.$$ Energy, however, is not conserved as electrostatic energy. Working out $\tfrac12C_1V_1^2+\tfrac12C_2V_2^2-\tfrac12(C_1+C_2)V^2$ gives a loss $$\Delta U=\frac{C_1C_2}{2(C_1+C_2)}\,(V_1-V_2)^2,$$ which is never negative and is zero only if the voltages were already equal. The lost energy is dissipated as heat in the connecting wires (and a little as electromagnetic radiation) while the charge surges across — however small the resistance of the wires.

📐 Result 2.16 — Common potential and the energy lost in sharing

When capacitors $C_1$ at $V_1$ and $C_2$ at $V_2$ are joined like plate to like plate, charge is conserved and they reach the common potential $V=\dfrac{C_1V_1+C_2V_2}{C_1+C_2}$. The electrostatic energy decreases by $\Delta U=\dfrac{C_1C_2(V_1-V_2)^2}{2(C_1+C_2)}$. (If they are joined positive plate to negative plate, replace $V_2$ by $-V_2$.)

✏️ Example 6 — Sharing charge, losing energy
A $6.0\ \mu\text{F}$ capacitor is charged to $150$ V, disconnected from the supply, and then connected across an uncharged $3.0\ \mu\text{F}$ capacitor. Find (a) the common potential, (b) the final charge on each, and (c) the energy before and after, and the energy lost.
(a) Common potential Charge $Q=C_1V_1=(6.0\ \mu\text{F})(150\ \text{V})=900\ \mu\text{C}$ is shared: $V=\dfrac{900\ \mu\text{C}}{9.0\ \mu\text{F}}=100$ V.
(b) Charges $Q_1=6.0\times100=600\ \mu\text{C}$, $Q_2=3.0\times100=300\ \mu\text{C}$; total $900\ \mu\text{C}$ ✓.
(c) Energies Before: $\tfrac12(6.0\times10^{-6})(150)^2=6.75\times10^{-2}$ J. After: $\tfrac12(9.0\times10^{-6})(100)^2=4.5\times10^{-2}$ J. Lost: $2.25\times10^{-2}$ J — a third of the original energy.
Check $\Delta U=\dfrac{(6.0)(3.0)}{2(9.0)}\times(150)^2\times10^{-6}=1.0\times22500\times10^{-6}=2.25\times10^{-2}$ J ✓. Interactive 2 opens on this example.

Energy and dielectrics. The energy formulas also settle what happens when a dielectric slab ($K$) is pushed into a capacitor. If the capacitor is isolated, $Q$ is fixed and $U=Q^2/2C$ falls to $U_0/K$: the field does positive work pulling the slab in. If the battery stays connected, $V$ is fixed and $U=\tfrac12CV^2$ rises to $KU_0$; the battery supplies twice that increase ($\Delta Q\cdot V$), and the other half again goes into pulling the slab in.

⚠️ Common mistakes with combinations and energy

1. Swapping the rules with resistors. Capacitors in parallel add directly; in series their reciprocals add — the opposite of resistors.

2. Giving series capacitors the same voltage. In series the charge is the same; the voltage divides, most across the smallest capacitor.

3. Writing $U=QV$. The stored energy is $\tfrac12QV$. The battery does supply $QV$ while charging — the other half is dissipated as heat in the circuit.

4. Assuming energy is conserved when capacitors share charge. Charge is conserved; electrostatic energy is not (Result 2.16).

5. Using the wrong form of $U$. If $Q$ is fixed use $Q^2/2C$; if $V$ is fixed use $\tfrac12CV^2$. Using the other one gives the opposite trend when $C$ changes.

Interactive 2 — Sharing Charge and Losing Energy

$C_1$ is charged to $V_1$ and $C_2$ to $V_2$. Press Play to close the switch: charge surges through the wire (the resistance $R$ of the wire only sets how fast) until both reach the common potential. The graph tracks both voltages; the bars compare total charge (conserved) and total energy (not conserved) with their starting values.

6.03.01500
$V_1$, $V_2$ now—
common potential—
total charge—
energy: start → now—
🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — the force between the plates

Not required by the CBSE syllabus. The energy stored in a capacitor is listed by JEE Main, NEET and JEE Advanced; the force between the plates follows from it, although no syllabus names it.

The plates of a charged capacitor attract each other. To find the force, imagine pulling the plates of an isolated capacitor apart by a small extra distance $\Delta x$. $Q$ stays the same, so $E=\sigma/\varepsilon_0$ is unchanged, and the field simply fills an extra volume $A\,\Delta x$. The work done must equal the extra energy: $F\,\Delta x=\tfrac12\varepsilon_0E^2\,A\,\Delta x$, so $$F=\frac12\varepsilon_0E^2A=\frac{Q^2}{2\varepsilon_0A}=\frac12QE .$$ Note the $\tfrac12$: each plate feels the field of the other plate only, which is $E/2$. The same energy method shows that a dielectric slab is pulled into a capacitor, which is used in some sensors and in JEE problems on a slab oscillating in and out of a capacitor.

Exercises

I
Series and parallel combinations.
  • 1 Find the equivalent capacitance of $2.0\ \mu\text{F}$ and $3.0\ \mu\text{F}$ (a) in series, (b) in parallel.
  • 2 Three $6.0\ \mu\text{F}$ capacitors are connected (a) in series, (b) in parallel. Find the equivalent capacitance in each case.
  • 3 $3.0\ \mu\text{F}$ and $6.0\ \mu\text{F}$ capacitors in series are connected to $18$ V. Find the charge on each and the voltage across each.
  • 4 The same two capacitors are connected in parallel to $18$ V. Find the charge on each.
  • 5 How many $8.0\ \mu\text{F}$ capacitors must be connected in series to give $2.0\ \mu\text{F}$?
  • 6 A $2.0\ \mu\text{F}$ capacitor in parallel with a $4.0\ \mu\text{F}$ capacitor is connected in series with a $3.0\ \mu\text{F}$ capacitor, and the network is put across $12$ V. Find the equivalent capacitance, and the charge on and voltage across each capacitor.
  • 7 Two $4.0\ \mu\text{F}$ capacitors in series are connected in parallel with a $2.0\ \mu\text{F}$ capacitor, across $10$ V. Find the equivalent capacitance and the charge on each capacitor.
  • 8 $n$ identical capacitors $C$ are connected first in series and then in parallel. Find the ratio of the parallel to the series equivalent capacitance.
Check: series and parallel equivalent of up to four capacitors
series $C$; $Q$ on each—
parallel $C$; total $Q$—
energy: series, parallel—
1 (a) $\dfrac{(2.0)(3.0)}{5.0}=1.2\ \mu\text{F}$; (b) $5.0\ \mu\text{F}$.
2 (a) $\dfrac{6.0}{3}=2.0\ \mu\text{F}$; (b) $18\ \mu\text{F}$.
3 $C=2.0\ \mu\text{F}$, $Q=36\ \mu\text{C}$ on each. $V_3=\dfrac{36}{3.0}=12$ V, $V_6=\dfrac{36}{6.0}=6.0$ V.
4 $18$ V across each: $Q_3=54\ \mu\text{C}$, $Q_6=108\ \mu\text{C}$.
5 $n$ in series give $\dfrac{8.0}{n}=2.0\Rightarrow n=4$.
6 Pair: $6.0\ \mu\text{F}$; with $3.0\ \mu\text{F}$ in series: $C=\dfrac{(6.0)(3.0)}{9.0}=2.0\ \mu\text{F}$. $Q=24\ \mu\text{C}$ on the $3.0\ \mu\text{F}$ (so $8.0$ V) and on the pair (so $4.0$ V). $Q_2=2.0\times4.0=8.0\ \mu\text{C}$, $Q_4=16\ \mu\text{C}$.
7 Series pair: $2.0\ \mu\text{F}$; plus $2.0\ \mu\text{F}$ in parallel: $C=4.0\ \mu\text{F}$. Each branch has $10$ V: the $2.0\ \mu\text{F}$ capacitor holds $20\ \mu\text{C}$; the series pair holds $20\ \mu\text{C}$, so each $4.0\ \mu\text{F}$ capacitor carries $20\ \mu\text{C}$ with $5.0$ V across it. Total $40\ \mu\text{C}$.
8 $C_s=C/n$, $C_p=nC$, ratio $n^2$.
II
Energy stored. $\varepsilon_0=8.85\times10^{-12}$ F/m.
  • 1 Find the energy stored in a $10\ \mu\text{F}$ capacitor charged to $50$ V.
  • 2 A capacitor stores $0.50$ J when charged to $100$ V. Find its capacitance.
  • 3 A $2.0\ \mu\text{F}$ capacitor carries $40\ \mu\text{C}$. Find the energy stored.
  • 4 The voltage across a capacitor is doubled. By what factor does the stored energy change? What if instead the charge is doubled?
  • 5 Find the energy density of the electric field in air just below breakdown, $E=3.0\times10^6$ V/m.
  • 6 A parallel-plate capacitor with $A=0.020\ \text{m}^2$ and $d=1.0$ mm (air) is charged to $500$ V. Find the stored energy from $\tfrac12CV^2$ and from the energy density, and show that they agree.
  • 7 $2.0\ \mu\text{F}$ and $3.0\ \mu\text{F}$ capacitors in series are connected to $100$ V. Find the energy stored in each and check the total against $\tfrac12C_\text{eq}V^2$.
  • 8 Repeat part 7 with the capacitors in parallel. Which arrangement stores more energy, and why?
1 $U=\tfrac12(10\times10^{-6})(50)^2=1.25\times10^{-2}$ J.
2 $C=\dfrac{2U}{V^2}=\dfrac{1.0}{10^4}=1.0\times10^{-4}$ F $=100\ \mu\text{F}$.
3 $U=\dfrac{Q^2}{2C}=\dfrac{(40\times10^{-6})^2}{2(2.0\times10^{-6})}=4.0\times10^{-4}$ J.
4 $U=\tfrac12CV^2$: doubling $V$ (same capacitor) multiplies $U$ by $4$. Doubling $Q$ also doubles $V$ (since $C$ is fixed), so again $\times4$.
5 $u=\tfrac12(8.85\times10^{-12})(3.0\times10^6)^2\approx40\ \text{J/m}^3$ — the most energy an air capacitor can hold per cubic metre, which is why practical capacitors use dielectrics with high breakdown fields.
6 $C=\dfrac{(8.85\times10^{-12})(0.020)}{1.0\times10^{-3}}=1.77\times10^{-10}$ F; $U=\tfrac12(1.77\times10^{-10})(500)^2\approx2.2\times10^{-5}$ J. Field $E=5.0\times10^5$ V/m, $u=\tfrac12\varepsilon_0E^2\approx1.11\ \text{J/m}^3$, volume $2.0\times10^{-5}\ \text{m}^3$, product $\approx2.2\times10^{-5}$ J ✓.
7 $C_\text{eq}=1.2\ \mu\text{F}$, $Q=120\ \mu\text{C}$. $U_2=\dfrac{Q^2}{2C_2}=\dfrac{1.44\times10^{-8}}{4.0\times10^{-6}}=3.6\times10^{-3}$ J; $U_3=\dfrac{1.44\times10^{-8}}{6.0\times10^{-6}}=2.4\times10^{-3}$ J. Total $6.0\times10^{-3}$ J $=\tfrac12(1.2\times10^{-6})(100)^2$ ✓. In series the smaller capacitor stores more.
8 $U_2=\tfrac12(2.0\times10^{-6})(100)^2=1.0\times10^{-2}$ J, $U_3=1.5\times10^{-2}$ J, total $2.5\times10^{-2}$ J. Parallel stores more: the same voltage across a larger equivalent capacitance ($5.0$ vs $1.2\ \mu\text{F}$).
III
Sharing charge, and energy with dielectrics.
  • 1 A $5.0\ \mu\text{F}$ capacitor charged to $120$ V is connected across an identical uncharged capacitor. Find the common potential and the energy before and after.
  • 2 A $2.0\ \mu\text{F}$ capacitor at $100$ V and a $3.0\ \mu\text{F}$ capacitor at $50$ V are joined positive plate to positive plate. Find the common potential and the energy lost.
  • 3 Repeat part 2 with the positive plate of each joined to the negative plate of the other. Find the common potential and the final energy.
  • 4 In part 1 half of the energy disappears, however small the resistance of the connecting wires. Where does it go?
  • 5 A $200$ pF air capacitor is charged to $100$ V and isolated. A slab with $K=5.0$ is then pushed in to fill it. Find the energy before and after. Is the slab pulled in or must it be pushed?
  • 6 Repeat part 5 with the battery left connected. Find the change in stored energy and the energy supplied by the battery.
  • 7 A $2.0\ \mu\text{F}$ capacitor at $200$ V is connected across an uncharged $6.0\ \mu\text{F}$ capacitor. Find the final charge on each.
1 $V=\dfrac{(5.0)(120)}{10}=60$ V. Before: $\tfrac12(5.0\times10^{-6})(120)^2=3.6\times10^{-2}$ J. After: $\tfrac12(10\times10^{-6})(60)^2=1.8\times10^{-2}$ J — half is lost.
2 $V=\dfrac{(2.0)(100)+(3.0)(50)}{5.0}=\dfrac{350}{5.0}=70$ V. $\Delta U=\dfrac{(2.0)(3.0)}{2(5.0)}(100-50)^2\times10^{-6}=1.5\times10^{-3}$ J. (Before $1.375\times10^{-2}$ J, after $1.225\times10^{-2}$ J.)
3 Net charge $200-150=50\ \mu\text{C}$, so $V=\dfrac{50}{5.0}=10$ V; final energy $\tfrac12(5.0\times10^{-6})(10)^2=2.5\times10^{-4}$ J — most of the energy is lost.
4 Into heat in the wires (the current is larger when $R$ is smaller, and the heat $\int I^2R\,dt$ comes out the same) and a little into electromagnetic radiation from the sudden surge of current.
5 $Q=CV=2.0\times10^{-8}$ C. Before: $\tfrac12CV^2=1.0\times10^{-6}$ J. After: $\dfrac{Q^2}{2KC}=\dfrac{1.0\times10^{-6}}{5.0}=2.0\times10^{-7}$ J. The energy falls, so the field does work on the slab: it is pulled in.
6 $V$ fixed: $U=\tfrac12(KC)V^2=5.0\times10^{-6}$ J, an increase of $4.0\times10^{-6}$ J. The battery moves extra charge $\Delta Q=(KC-C)V=8.0\times10^{-8}$ C and supplies $\Delta Q\,V=8.0\times10^{-6}$ J — twice the increase. The other $4.0\times10^{-6}$ J is the work done in pulling the slab in (and is finally dissipated).
7 $V=\dfrac{(2.0)(200)}{8.0}=50$ V; $Q_1=100\ \mu\text{C}$, $Q_2=300\ \mu\text{C}$.
IV
Capacitors at work.
  • 1 Modelling. A camera flash uses a $220\ \mu\text{F}$ capacitor charged to $330$ V. Find the energy stored, and the average power if it is released in $1.0$ ms.
  • 2 Modelling. What capacitance must a defibrillator have to store $360$ J at $3.0$ kV?
  • 3 Modelling. A supercapacitor is rated $3000$ F at $2.7$ V. Find the energy it stores and compare it with an AA cell (about $1\times10^{4}$ J).
  • 4 Modelling. Below a thundercloud the field is about $1.0\times10^5$ V/m over a region $1.0\ \text{km}\times1.0$ km in area and $2.0$ km high. Estimate the energy stored in this field.
  • 5 A $2.0\ \mu\text{F}$ and a $4.0\ \mu\text{F}$ capacitor, each rated at $100$ V, are connected in series. What is the greatest voltage that can be applied across the combination?
  • 6 Explain why connecting capacitors in series raises the working voltage of the combination but lowers its capacitance.
1 $U=\tfrac12(220\times10^{-6})(330)^2\approx12$ J. $P=\dfrac{12}{1.0\times10^{-3}}\approx1.2\times10^4$ W.
2 $C=\dfrac{2U}{V^2}=\dfrac{720}{9.0\times10^6}=8.0\times10^{-5}$ F $=80\ \mu\text{F}$.
3 $U=\tfrac12(3000)(2.7)^2\approx1.1\times10^{4}$ J — about the same as an AA cell, but deliverable in seconds and over a million charge cycles.
4 $u=\tfrac12(8.85\times10^{-12})(1.0\times10^5)^2\approx4.4\times10^{-2}\ \text{J/m}^3$; volume $(1.0\times10^3)^2(2.0\times10^3)=2.0\times10^9\ \text{m}^3$; $U\approx8.9\times10^{7}$ J — tens of megajoules, of the same order as the energy of a lightning flash.
5 Same charge on both, so $V\propto1/C$: the $2.0\ \mu\text{F}$ capacitor takes twice the voltage of the $4.0\ \mu\text{F}$ one. It reaches its $100$ V limit when the other has $50$ V: $V_\text{max}=150$ V.
6 In series the applied voltage is shared, so each capacitor sees only part of it and the total can exceed one rating. But the same charge $Q$ now needs the whole voltage $V$, so $C=Q/V$ is smaller — like one capacitor with a wider gap.

Original problems in the style of the CBSE board paper, JEE Main, JEE Advanced and NEET. Try each one against the clock (about 2 minutes for an MCQ, 5 minutes for a long answer) before opening the solution.

🎯 Exam corner — combinations and energy BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks Derive an expression for the energy stored in a charged capacitor, and show that for a parallel-plate capacitor the energy per unit volume of the field is $\tfrac12\varepsilon_0E^2$. Find the energy stored in a $44$ pF air capacitor charged to $100$ V.

Work At charge $q$ the voltage is $q/C$; moving $dq$ more needs $dW=\dfrac qC\,dq$. Total $W=\displaystyle\int_0^Q\frac qC\,dq=\frac{Q^2}{2C}=\tfrac12CV^2$, stored as $U$.
Density With $C=\dfrac{\varepsilon_0A}{d}$ and $V=Ed$: $U=\tfrac12\dfrac{\varepsilon_0A}{d}E^2d^2=\tfrac12\varepsilon_0E^2\,(Ad)$. Dividing by the volume $Ad$ of the field: $u=\tfrac12\varepsilon_0E^2$.
Number $U=\tfrac12(44\times10^{-12})(100)^2=2.2\times10^{-7}$ J. Exam tip: mention that the field outside the plates is zero — that is why only the volume $Ad$ counts.

E2. Board · 2 marks Three capacitors, each of $6.0\ \mu\text{F}$, are connected so that one is in series with a parallel combination of the other two. The network is connected to a $9.0$ V battery. Find the equivalent capacitance and the charge on each capacitor.

Reduce Parallel pair $12\ \mu\text{F}$; with $6.0\ \mu\text{F}$ in series: $C=\dfrac{(6.0)(12)}{18}=4.0\ \mu\text{F}$.
Charges $Q=CV=36\ \mu\text{C}$ on the single capacitor ($6.0$ V across it). The pair has $3.0$ V, so each of the two carries $18\ \mu\text{C}$. Exam tip: state the series/parallel reason at each step — "same charge in series" earns the method mark.

E3. JEE Main (Numerical value.) A $10\ \mu\text{F}$ capacitor is charged to $50$ V and then connected across an uncharged $15\ \mu\text{F}$ capacitor. Find the heat produced in the connecting wires, in mJ.

Formula $\Delta U=\dfrac{C_1C_2}{2(C_1+C_2)}(V_1-V_2)^2=\dfrac{(10)(15)}{2(25)}\times10^{-6}\times(50)^2$.
Answer $\Delta U=3.0\times10^{-6}\times2500=7.5\times10^{-3}$ J $=\mathbf{7.5}$ mJ. (Common potential $20$ V; energy $12.5$ mJ before, $5.0$ mJ after.) Exam tip: the heat does not depend on the resistance of the wires.

E4. NEET Capacitors $2C$ and $C$ are connected in series across a battery. The ratio of the energy stored in $2C$ to that stored in $C$ is (A) $1:2$ (B) $2:1$ (C) $1:4$ (D) $4:1$.

Concept Series: same $Q$, so $U=\dfrac{Q^2}{2C}\propto\dfrac1C$.
Answer (A) $\dfrac{U_{2C}}{U_C}=\dfrac{C}{2C}=\dfrac12$. Exam tip: in parallel (same $V$) the ratio would flip to $2:1$ — check which quantity is shared before choosing the formula.

E5. JEE Advanced (One or more correct.) An isolated air-filled parallel-plate capacitor carries charge $Q$. Its plates are pulled apart slowly until the separation is doubled. Which statements are correct? (A) The potential difference doubles. (B) The field between the plates is unchanged. (C) The stored energy doubles. (D) The energy density between the plates doubles.

Setup Isolated: $Q$ fixed; $C=\varepsilon_0A/d$ halves.
Options (A) $V=Q/C$ doubles ✓. (B) $E=\sigma/\varepsilon_0=Q/(\varepsilon_0A)$ is unchanged ✓. (C) $U=Q^2/2C$ doubles ✓ — the extra energy is the work done pulling the attracting plates apart. (D) $u=\tfrac12\varepsilon_0E^2$ is unchanged; the volume doubled ✗.
Answer (A), (B), (C). Exam tip: for an isolated capacitor think "same $Q$, same $E$"; for a connected one think "same $V$".

E6. Board · Case study A portable defibrillator charges a $100\ \mu\text{F}$ capacitor to $2.0$ kV and then discharges it through the patient. (a) Find the charge stored and the energy stored. (2 marks) (b) If a dielectric of twice the dielectric constant were used in a capacitor of the same dimensions, to what voltage would it need to be charged to store the same energy? (1 mark) (c) The discharge lasts $4.0$ ms. Find the average power delivered. (1 mark)

(a) $Q=CV=(100\times10^{-6})(2.0\times10^3)=0.20$ C; $U=\tfrac12CV^2=\tfrac12(1.0\times10^{-4})(4.0\times10^6)=200$ J.
(b) $C$ doubles; $\tfrac12(2C)V'^2=\tfrac12CV^2\Rightarrow V'=\dfrac{V}{\sqrt2}\approx1.4$ kV.
(c) $P=\dfrac{200}{4.0\times10^{-3}}=5.0\times10^4$ W. Exam tip: in case studies keep units on every line — "200 J" and "50 kW" are each worth a mark.
🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. Could a city bus run on capacitors alone, recharging for a few seconds at every stop — and how many capacitor modules would it need?

Why it matters. Since 2010 "capabuses" have carried passengers in Shanghai, and similar systems have been trialled in Europe: supercapacitors survive a million charge cycles, charge in seconds and work well in the cold, but store far less energy per kilogram than lithium-ion batteries. Deciding between them is a real engineering trade-off for transport planners in growing cities from Bengaluru to Bogotá.

What to do.

  1. Data: one supercapacitor module is rated $63$ F at $125$ V and has a mass of $60$ kg. The bus uses $1.0$ kWh of electrical energy per km, stops are $0.50$ km apart, and the charger at each stop delivers $200$ kW.
  2. Find the energy stored in one fully charged module, in J and in kWh.
  3. The motor controller cannot use the module below half its rated voltage. What fraction of the stored energy is usable? Find the usable energy per module.
  4. Find the energy needed between stops, the number of modules required and their total mass, and the charging time at each stop.
  5. A lithium-ion pack stores about $5\times10^{5}$ J/kg. What mass of batteries would hold the same usable energy? Discuss the advantages of each choice.

Tools: Calculator; $1\ \text{kWh}=3.6\times10^6$ J.

Hand in: A one-page engineering note with your calculations, a table comparing the two options (mass, charging time, lifetime) and a recommendation.

Going further. Real buses also recover energy when braking. If the bus (mass $1.5\times10^4$ kg) stops from $40$ km/h at every stop and $60\%$ of its kinetic energy is returned to the capacitors, what fraction of the energy needed per stop does braking supply?

Instructor note: one module $U=\tfrac12(63)(125)^2\approx4.9\times10^5$ J $\approx0.14$ kWh. Down to half voltage the usable fraction is $1-\tfrac14=\tfrac34$, about $3.7\times10^5$ J. Between stops $0.50$ kWh $=1.8\times10^6$ J, so $4.9\to5$ modules ($300$ kg); charging time $1.8\times10^6/2\times10^5=9$ s. Batteries: $1.8\times10^6/5\times10^5\approx3.6$ kg of cells for the same usable energy — capacitors win on cycle life and charging speed, batteries on range. Braking: $\tfrac12(1.5\times10^4)(11.1)^2\approx9.3\times10^5$ J, $60\%$ of which is $5.6\times10^5$ J, about $31\%$ of the energy per stop.

🧪 Self-check quiz

0/5 correct

Q1. The equivalent capacitance of $3\ \mu\text{F}$ and $6\ \mu\text{F}$ in series is

Q2. Unequal capacitors connected in series across a battery all have the same

Q3. The potential difference across a given capacitor is doubled. The energy stored becomes

Q4. The energy stored per unit volume in an electric field $E$ in vacuum is

Q5. A charged capacitor is connected across an identical uncharged capacitor. The total stored energy becomes

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