JEE Mains Physics · Electrostatics

Energy Stored and Charge Sharing in Capacitors

A capacitor stores ½CV² = Q²/2C; first ask what stays fixed, V while the battery is connected or Q once it is removed, and when two capacitors share charge, the charge is kept but some energy is lost.

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Why this matters

Twenty-three PYQs, twelve of them multiple choice and eleven asking for a number, and two from 2026. Eight find the stored energy and how it changes when a slab goes in, with the battery connected or removed. Fifteen join a charged capacitor to another one and ask for the common potential, the charge that moves or the energy lost.

Concept 1 of 2: Stored energy: battery connected or removed

Charging a capacitor is work done against the growing voltage, and it is stored as ½CV². When something changes C, the answer depends on what is held fixed. With the battery connected, V stays the same, so more C means more charge and more energy. With the battery removed, Q is trapped on the plates, so more C means less voltage and less energy.

Definition

$U = \tfrac{1}{2}CV^{2} = \dfrac{Q^{2}}{2C} = \tfrac{1}{2}QV$ . Energy per unit volume: $\tfrac{1}{2}K\varepsilon_0E^{2}$ .
Battery connected (V fixed), slab K fills the gap: C, Q and U all grow K times. Extra charge $(K - 1)CV$ , extra energy $\tfrac{1}{2}(K - 1)CV^{2}$ . The battery does $(K - 1)CV^{2}$ of work, twice the energy gained.
Battery removed (Q fixed), slab K fills the gap: V, E and U all fall K times. Energy lost $\tfrac{1}{2}CV^{2}\left(1 - \dfrac{1}{K}\right)$ ; the field pulls the slab in.
Plates pulled apart: with V fixed, U falls; with Q fixed, U grows.
The same supply across a group: $\dfrac{U_{\text{parallel}}}{U_{\text{series}}} = \dfrac{C_{\text{parallel}}}{C_{\text{series}}}$ .

Stored energy

U = \tfrac{1}{2}CV^{2} = \frac{Q^{2}}{2C}, \qquad V\ \text{fixed: } U \to KU, \qquad Q\ \text{fixed: } U \to \frac{U}{K}

Worked example

10\ \mu\text{F}

capacitor is charged to 300 V and the battery is removed. A slab with K = 4 then fills the gap. Find the new voltage and energy, and the change in energy.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 22 Jan 2025 · Q19Moderate

Example 1 · Electrostatics · Energy Stored and Charge Sharing in Capacitors

A parallel-plate capacitor of capacitance

40\mu\text{ }F

is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant

K = 2

. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are -

Decide what is fixed first

Battery connected: V fixed, use ½CV². Battery removed: Q fixed, use Q²/2C. Using the other form gives the change in the wrong direction.

The battery supplies twice the gain

With V fixed, the battery's work (K − 1)CV² is twice the rise in stored energy. The other half goes into pulling the slab in.

Percentages are squared

Energy goes as Q² or V². A 10% rise in charge is a 21% rise in energy, not 10%.

Drill 7 more on stored energy: battery connected or removed

Concept 2 of 2: Charge sharing and common potential

Join two charged capacitors and charge flows until both are at the same voltage. The total charge cannot change, so the common voltage is the total charge over the total capacitance. Energy, though, is not kept: some is lost as heat and radiation in the connecting wires while the charge flows. The loss is zero only if both were already at the same voltage.

Definition

Like plates joined (+ to +): $V = \dfrac{C_1V_1 + C_2V_2}{C_1 + C_2}$ . Unlike plates (+ to −): $V = \dfrac{|C_1V_1 - C_2V_2|}{C_1 + C_2}$ .
Charge after joining: $Q_i = C_iV$ on each.
Energy lost: $\Delta U = \dfrac{C_1C_2}{2(C_1 + C_2)}(V_1 - V_2)^{2}$ for like plates; use $(V_1 + V_2)^{2}$ for unlike plates.
Identical capacitors, one uncharged: the voltage halves and half the stored energy is lost.
A slab put into one of two joined capacitors after the battery is removed: the total Q stays, so the new common V is $\dfrac{Q_{\text{total}}}{KC_1 + C_2}$ .
Spheres joined by a wire follow the same rule, with $C = 4\pi\varepsilon_0 R$ .

Common potential and energy lost

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}

Worked example

4\ \mu\text{F}

capacitor at 100 V is joined, like plates together, to a

6\ \mu\text{F}

capacitor at 50 V. Find the common voltage, the final charges and the energy lost.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 1 February 2024 · Q5Moderate

Example 2 · Electrostatics · Energy Stored and Charge Sharing in Capacitors

Two identical capacitors have same capacitance

C

. One of them is charged to the potential

V

and other to the potential

2\text{ }V

. The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is:

Unlike plates subtract

Joining positive to negative cancels part of the charge first. Adding C₁V₁ and C₂V₂ here gives a common voltage that is far too high.

Charge is kept, energy is not

The final energy is always less, unless the two voltages were already equal. Setting the energies equal before and after gives a wrong voltage.

Each capacitor's share is CV

After joining, each capacitor holds its own C times the common V. The charge splits equally only when the capacitances are equal.

Drill 14 more on charge sharing and common potential

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 (2)

Stored energy: battery connected or removed
Stored energy
$U = \tfrac{1}{2}CV^{2} = \frac{Q^{2}}{2C}, \qquad V\ \text{fixed: } U \to KU, \qquad Q\ \text{fixed: } U \to \frac{U}{K}$
Charge sharing and common potential
Common potential and energy lost
$V = \frac{C_1V_1 + C_2V_2}{C_1 + C_2}, \qquad \Delta U = \frac{C_1C_2}{2(C_1 + C_2)}(V_1 - V_2)^{2}$

Watch out for (6)

Decide what is fixed first
→ Stored energy: battery connected or removed
The battery supplies twice the gain
→ Stored energy: battery connected or removed
Percentages are squared
→ Stored energy: battery connected or removed
Unlike plates subtract
→ Charge sharing and common potential
Charge is kept, energy is not
→ Charge sharing and common potential
Each capacitor's share is CV
→ Charge sharing and common potential

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