MHT-CET Physics · Electrostatics

Energy Stored in a Capacitor

A charged capacitor stores U = ½CV² = Q²/2C in the field between its plates; joining capacitors, re-arranging them or pulling their plates apart moves that energy around, and each change is found by comparing U before and after.

All Electrostatics notes MHT-CET Physics strategy

Why this matters

21 PYQs, six HARD — three of them two charged capacitors joined together, and the energy lost as their charge redistributes. The rest ask for U in one of its three forms, compare the energy of series and parallel groups, or ask what pulling an isolated capacitor's plates apart does to its voltage and how much work it takes.

Concept 1 of 4: Three Forms of the Stored Energy

Charging a capacitor pushes each new bit of charge against the voltage already there, which rises from 0 to V — so the average push is V/2 and the energy is ½QV. Swap in Q = CV to get the other two forms. Spread over the volume between the plates, it is an energy density ½ε₀E².

Definition

$U = \dfrac{1}{2}CV^2 = \dfrac{Q^2}{2C} = \dfrac{1}{2}QV$ .
Energy density between the plates: $u = \dfrac{1}{2}\varepsilon_0E^2 = \dfrac{\sigma^2}{2\varepsilon_0} = \dfrac{q^2}{2\varepsilon_0A^2}$ ; total $U = u \times Ad$ .
$U \propto Q^2$ : charge up by 20% means energy up by 44%; charge up by 10%, energy up by 21%.
Work to raise the voltage from $V_1$ to $V_2$ : $\dfrac{1}{2}C(V_2^2 - V_1^2)$ .
All the energy of $C_1$ at $V_1$ moved into $C_2$ : $C_2V_2^2 = C_1V_1^2$ .
Charged in parallel on one battery, $C_1$ and $C_2$ : $\dfrac{E_1 - E_2}{Q_1 - Q_2} = \dfrac{V}{2}$ .

Stored energy

U = \tfrac{1}{2}CV^2 = \frac{Q^2}{2C} = \tfrac{1}{2}QV, \qquad u = \tfrac{1}{2}\varepsilon_0E^2

Worked example

20\,\mu

F capacitor is charged to 50 V. Energy stored? And the energy density in a field of

10^5

V/m?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 2 · Q49Moderate

Example 1 · Electrostatics · Energy Stored in a Capacitor

If the charge on the capacitor is increased by 3 coulomb, the energy stored in it increases by 44%. The original charge on the capacitor is

Treating energy as proportional to charge

Energy goes as

Q^2

(or

V^2

). A 44% rise in energy is a 20% rise in charge; the 3 C added is then 20% of the original, 15 C.

Drill 6 more on three forms of the stored energy

Concept 2 of 4: Energy of Series and Parallel Groups

A group stores the energy of its equivalent capacitor: ½C_eq V². For the same energy, a small C_eq needs a large voltage. Re-arranging charged capacitors without a battery moves no charge off them, so the energy stays the same.

Definition

Group energy $U = \dfrac{1}{2}C_{\text{eq}}V^2$ . $n$ identical in series: $\dfrac{C}{n}$ ; in parallel: $nC$ .
Same energy in series and in parallel: $\dfrac{V_s}{V_p} = \sqrt{\dfrac{C_p}{C_s}}$ ; for $n$ identical, $V_s : V_p = n : 1$ .
$n$ capacitors charged in parallel to $V$ , then separated and joined in series: total voltage $nV$ , energy unchanged.
$n_1$ of $C_1$ in series at $V_1$ versus $n_2$ of $C_2$ in parallel at $V_2$ , equal energy: $\dfrac{C_1}{n_1}V_1^2 = n_2C_2V_2^2$ .

Equal energy, series and parallel

\tfrac{1}{2}C_sV_s^2 = \tfrac{1}{2}C_pV_p^2 \;\Rightarrow\; \frac{V_s}{V_p} = \sqrt{\frac{C_p}{C_s}}

Worked example

Three identical capacitors must store the same energy connected in series as in parallel. Ratio of the voltages needed?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 2 · Q29Moderate

Example 2 · Electrostatics · Energy Stored in a Capacitor

The potential difference that must be applied across the series and parallel combination of 4 identical capacitors is such that the energy stored in them becomes the same. The ratio of potential difference in series to parallel combination is

Squaring the capacitance ratio

Equal energy gives

\frac{V_s^2}{V_p^2} = \frac{C_p}{C_s}

, so the VOLTAGE ratio is the square root. For four identical capacitors that is 4 : 1, while 16 : 1 is the ratio of the capacitances.

Drill 5 more on energy of series and parallel groups

Concept 4 of 4: Pulling the Plates Apart: Work and Force

Pull the plates of an isolated capacitor apart and the charge stays; the capacitance falls, so the voltage and the energy both rise in proportion to the gap. The extra energy is the work you did against the plates' attraction. The attraction itself is constant, Q²/2ε₀A, because the field between the plates does not change.

Definition

Isolated (Q fixed), gap $d \to nd$ : $C \to \dfrac{C}{n}$ , $V \to nV$ , $E$ unchanged, $U \to nU$ . Work done $= (n - 1)U_i = (n - 1)\dfrac{\varepsilon_0AV^2}{2d}$ .
Force between the plates: $F = \dfrac{Q^2}{2\varepsilon_0A} = \dfrac{CV^2}{2d}$ — each plate feels the field of the OTHER, $\frac{E}{2}$ .
With the battery connected instead (V fixed), widening the gap LOWERS $Q$ , $E$ and $U$ .

Isolated capacitor, gap multiplied by n

W = (n - 1)\,\frac{\varepsilon_0AV^2}{2d}, \qquad F = \frac{Q^2}{2\varepsilon_0A}

Worked example

A 10 pF capacitor is charged to 100 V and isolated. Its gap is doubled. Work done?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 21 April Shift II · Q38Moderate

Example 4 · Electrostatics · Energy Stored in a Capacitor

A parallel plate capacitor having plate area '

A

' and separation '

d

' is charged to a potential difference '

V

'. The charging battery is disconnected, and the plates are pulled apart to four times the initial separation. The work required to increase the distance between the plates is (

\varepsilon_{0}=

permittivity of free space)

Taking the final energy as the work

The work is the INCREASE,

U_f - U_i

. Pulled to 4 times the gap,

U_f = 4U_i

but the work is

3U_i = \frac{3\varepsilon_0AV^2}{2d}

;

\frac{2\varepsilon_0AV^2}{d}

is the final energy and it is printed too.

Drill 4 more on pulling the plates apart: work and force

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

Three Forms of the Stored Energy
Stored energy
$U = \tfrac{1}{2}CV^2 = \frac{Q^2}{2C} = \tfrac{1}{2}QV, \qquad u = \tfrac{1}{2}\varepsilon_0E^2$
Energy of Series and Parallel Groups
Equal energy, series and parallel
$\tfrac{1}{2}C_sV_s^2 = \tfrac{1}{2}C_pV_p^2 \;\Rightarrow\; \frac{V_s}{V_p} = \sqrt{\frac{C_p}{C_s}}$
Joining Two Charged Capacitors: Common Potential and Energy Lost
Common potential and loss
$V = \frac{C_1V_1 + C_2V_2}{C_1 + C_2}, \qquad \Delta U = \frac{C_1C_2}{2(C_1 + C_2)}(V_1 - V_2)^2$
Pulling the Plates Apart: Work and Force
Isolated capacitor, gap multiplied by n
$W = (n - 1)\,\frac{\varepsilon_0AV^2}{2d}, \qquad F = \frac{Q^2}{2\varepsilon_0A}$

Watch out for (4)

Treating energy as proportional to charge
→ Three Forms of the Stored Energy
Squaring the capacitance ratio
→ Energy of Series and Parallel Groups
Energy conserved by assumption
→ Joining Two Charged Capacitors: Common Potential and Energy Lost
Taking the final energy as the work
→ Pulling the Plates Apart: Work and Force

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