JEE Mains Physics · Electrostatics

Capacitance and Capacitor Combinations

Capacitance C = Q/V depends only on shape, size and the medium; in series every capacitor holds the same charge and 1/C adds, in parallel every one has the same voltage and C adds.

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Why this matters

Thirty-three PYQs, twenty-three of them multiple choice, and four from 2026. Seventeen are about a single capacitor: the capacitance of plates and spheres, the field and energy between the plates, facts about dielectrics, leakage and charging current. Sixteen reduce a network: series and parallel, bridges and ladders, and capacitors inside DC circuits once the currents have settled.

Concept 1 of 2: Capacitance of plates and spheres

A capacitor stores charge at a potential difference, and C tells you how much charge per volt. Bigger plates hold more charge; closer plates let a small voltage make a strong field. Neither the charge nor the voltage changes C: double the charge and the voltage doubles with it. A dielectric between the plates weakens the field for the same charge, so C grows by K.

Definition

$C = \dfrac{Q}{V}$ . Parallel plates: $C = \dfrac{K\varepsilon_0 A}{d}$ .
Isolated sphere: $C = 4\pi\varepsilon_0 R$ , the same for hollow and solid spheres. Spherical capacitor with the outer sphere earthed: $\dfrac{4\pi\varepsilon_0 R_1R_2}{R_2 - R_1}$ .
Between the plates: $E = \dfrac{\sigma}{\varepsilon_0} = \dfrac{V}{d}$ ; energy per unit volume $\tfrac{1}{2}\varepsilon_0E^{2}$ ; force between the plates $\dfrac{Q^{2}}{2\varepsilon_0 A}$ .
Plates carrying $q_1$ and $q_2$ : the inner faces hold $\pm\dfrac{q_1 - q_2}{2}$ , so $V = \dfrac{q_1 - q_2}{2C}$ .
Dielectrics: the induced surface charge is $Q\left(1 - \dfrac{1}{K}\right)$ . Non-polar molecules have no permanent dipole; polar ones do, but at random, so a sample has none until a field lines them up. A material's breakdown field caps the voltage.
A leaky dielectric of resistivity ρ: $RC = \rho K\varepsilon_0$ , whatever the plate size. While charging, the current is $i = C\dfrac{dV}{dt}$ .

Plates and spheres

C = \frac{K\varepsilon_0 A}{d}, \qquad C_{\text{sphere}} = 4\pi\varepsilon_0 R, \qquad C_{\text{spherical}} = \frac{4\pi\varepsilon_0 R_1R_2}{R_2 - R_1}

Worked example

An air capacitor has plates of area 0.02 m² that are 1 mm apart, with 50 V across it. Find C, Q, the field between the plates and the energy per unit volume. (

\varepsilon_0 = 8.85 \times 10^{-12}

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 25 July 2022 · Q96Moderate

Example 1 · Electrostatics · Capacitance and Capacitor Combinations

Capacitance of an isolated conducting sphere of radius

R_{1}

becomes

n

times when it is enclosed by a concentric conducting sphere of radius

R_{2}

connected to earth. The ratio of their radii

\left( \frac{R_{2}}{R_{1}} \right)

is:

C does not depend on Q or V

Charging a capacitor further raises Q and V together. Only the geometry and the medium change C.

Hollow or solid, the same C

A conductor's charge sits on its surface, so a hollow sphere and a solid one of the same radius have the same capacitance.

Unequal plate charges

Only half the difference of the two charges sits on the inner faces. Using q₁ alone for Q in V = Q/C gives a voltage that is too large.

Drill 16 more on capacitance of plates and spheres

Concept 2 of 2: Series, parallel and networks

In series, the charge that leaves one plate arrives at the next, so every capacitor holds the same charge and the voltages add. In parallel, every capacitor sits across the same two points, so the voltages are equal and the charges add. A complicated drawing becomes simple once you name its nodes: points joined by plain wire are one node, and each capacitor sits between two of them.

Definition

Series: same Q; $\dfrac{1}{C} = \sum \dfrac{1}{C_i}$ . The voltage splits in inverse proportion to C, and the total is less than the smallest.
Parallel: same V; $C = \sum C_i$ . The charge splits in proportion to C.
Two equal capacitors: series C/2, parallel 2C, a ratio of 1 : 4.
Network: label the nodes, merge points joined by wire, drop any capacitor whose two plates are on one node (it holds no charge), then combine step by step.
Bridge with $C_1/C_2 = C_3/C_4$ : the middle capacitor holds no charge.
DC circuit at steady state: no current flows through a capacitor's branch. Find the voltage across it from the resistor currents, then $Q = CV$ .

Series and parallel

\frac{1}{C_{s}} = \sum_i \frac{1}{C_i}, \qquad C_{p} = \sum_i C_i

Worked example

Capacitors of

3\ \mu\text{F}

and

6\ \mu\text{F}

in series are joined in parallel with a

4\ \mu\text{F}

capacitor, all across 12 V. Find the charge on each.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 11 April 2023 · Q111Moderate

Example 2 · Electrostatics · Capacitance and Capacitor Combinations

In the given circuit,

C_{1}= 2\mu F,C_{2}= 0.2\mu F,C_{3}= 2\mu F,C_{4}= 4\mu F

C_{5}= 2\mu F,C_{6}= 2\mu F

, the charge stored on capacitor

C_{4}

is___

\mu C

Decide by the nodes, not by the drawing

Two capacitors drawn one after the other are in parallel if both connect the same two nodes. Label the nodes before calling anything series.

A shorted capacitor stores nothing

If a wire joins a capacitor's two plates, both are on one node, so it holds no charge and drops out of the network.

Steady state means no capacitor current

Once the currents settle, a capacitor branch carries none. Remove the capacitors, find the currents, then read each capacitor's voltage from its two nodes.

Drill 15 more on series, parallel and networks

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)

Capacitance of plates and spheres
Plates and spheres
$C = \frac{K\varepsilon_0 A}{d}, \qquad C_{\text{sphere}} = 4\pi\varepsilon_0 R, \qquad C_{\text{spherical}} = \frac{4\pi\varepsilon_0 R_1R_2}{R_2 - R_1}$
Series, parallel and networks
Series and parallel
$\frac{1}{C_{s}} = \sum_i \frac{1}{C_i}, \qquad C_{p} = \sum_i C_i$

Watch out for (6)

C does not depend on Q or V
→ Capacitance of plates and spheres
Hollow or solid, the same C
→ Capacitance of plates and spheres
Unequal plate charges
→ Capacitance of plates and spheres
Decide by the nodes, not by the drawing
→ Series, parallel and networks
A shorted capacitor stores nothing
→ Series, parallel and networks
Steady state means no capacitor current
→ Series, parallel and networks

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