Capacitors for JEE Main

Capacitor questions in JEE Main reward bookkeeping more than insight. The trap is almost always the same: something changes (a slab goes in, a battery is removed, two capacitors are joined), and you have to know which quantity stays fixed. Decide that first, and the arithmetic is two lines.

The official 2026 syllabus places capacitors at the end of electrostatics: conductors and insulators, dielectrics and polarisation, capacitance, series and parallel combinations, the parallel-plate capacitor with and without a dielectric, and the energy stored. Revise electrostatics first if Gauss's law feels shaky, because the parallel-plate result comes straight from it.

Capacitance and the parallel-plate capacitor

A capacitor is two conductors carrying equal and opposite charges ±Q. Its capacitance is C = Q/V, measured in farads. C depends only on the geometry and the medium, not on Q or V.

Derivation. Plates of area A are a distance d apart, with charge density σ = Q/A. Between the plates the field is E = σ/ε₀ (the two sheets' fields add inside and cancel outside). So V = Ed = Qd/(ε₀A), and

C = ε₀A/d

An isolated sphere of radius R has C = 4πε₀R. Even a sphere the size of the Earth has only about 711 μF, which shows how large one farad is.

Dielectrics and slabs

A dielectric placed in the field is polarised, and its bound surface charges weaken the field inside by the factor K. With the whole gap filled, C = Kε₀A/d, K times the air value.

For slabs that fill only part of the gap, treat the pieces as capacitors in series or parallel:

A metal slab behaves like a dielectric with K infinitely large: the field inside it is zero.

Combinations

In series, the smallest capacitor takes the largest share of the voltage. That matters when each capacitor has a breakdown voltage: the small one fails first.

Energy stored

Charging from q to q + dq against a potential q/C needs dW = (q/C) dq. Adding up from 0 to Q:

U = Q²/(2C) = ½CV² = ½QV

The energy sits in the field, with density u = ½ε₀E² per unit volume. The attractive force between the plates is F = Q²/(2ε₀A).

When a battery charges a capacitor, it does work QV, but only ½QV is stored. The other half is lost as heat in the connecting wires, whatever their resistance.

Inserting a dielectric: battery on or off

Charge sharing

Join a charged capacitor to another one, positive plate to positive plate. Charge is conserved and the two end at a common potential:

V = (C₁V₁ + C₂V₂)/(C₁ + C₂)

Energy is not conserved. The loss is

ΔU = C₁C₂(V₁ − V₂)² / [2(C₁ + C₂)]

and it goes as heat and radiation during the flow of charge. If the plates are joined positive to negative, use C₁V₁ − C₂V₂ for the total charge, and V₁ + V₂ in place of V₁ − V₂ in the loss formula.

Worked problems

Problem 1: series combination. Capacitors of 4 μF and 12 μF are joined in series across 16 V. Find the charge and the voltage on each.

1/C = 1/4 + 1/12 = 4/12, so C = 3 μF and Q = 3 × 16 = 48 μC on each.
V₄ = 48/4 = 12 V and V₁₂ = 48/12 = 4 V. Check: 12 + 4 = 16 V.

Problem 2: half-filled capacitor. An air capacitor has C₀ = 10 μF. Find its capacitance when a slab of K = 4 (a) fills half the gap over the whole area, and (b) fills the whole gap over half the area.

(a) Series pieces: C = ε₀A/(d/2 + d/8) = ε₀A/(5d/8) = 1.6 C₀ = 16 μF.
(b) Parallel pieces: C = C₀/2 + 4C₀/2 = 2.5 C₀ = 25 μF.

Problem 3: charge sharing. A 2 μF capacitor charged to 100 V is joined to an identical uncharged capacitor. Find the common voltage and the energy lost.

Charge = 200 μC, shared over 4 μF, so V = 50 V.
Before: ½ × 2 × 10⁻⁶ × 100² = 10 mJ. After: ½ × 4 × 10⁻⁶ × 50² = 5 mJ.
Loss = 5 mJ. The formula agrees: (2 × 2 × 10⁻¹² × 100²)/(2 × 4 × 10⁻⁶) = 5 × 10⁻³ J.

Problem 4 (numerical answer): slab with the battery removed. A 5 μF capacitor is charged to 200 V and disconnected. A slab of K = 2 is then slid in to fill the gap. Find the energy stored afterwards, in mJ.

Before: ½ × 5 × 10⁻⁶ × 200² = 0.1 J = 100 mJ. Q is fixed, so U falls by K: 50 mJ. The other 50 mJ is the work done by the field in pulling the slab in.

Practice set

1. What happens to the capacitance of a parallel-plate capacitor when the plate separation is halved?
2. Three 6 μF capacitors are joined (a) in series and (b) in parallel. Find the equivalent capacitance each time.
3. Find the energy stored in a 10 μF capacitor charged to 100 V.
4. With the battery still connected, a slab of K = 3 fills the gap. How do Q, V, E and U change?
5. Plates of area 0.02 m² are 1 mm apart in air (ε₀ = 8.85 × 10⁻¹² F/m). Find C.
6. A metal slab of thickness d/2 is placed between plates d apart. How does C change?
7. A 3 μF capacitor at 100 V and a 6 μF capacitor at 40 V are joined (a) positive to positive and (b) positive to negative. Find the common voltage each time.
8. Find the energy density in a field of 10⁶ V/m in air.

Answers

1. It doubles, since C ∝ 1/d.
2. (a) 2 μF; (b) 18 μF.
3. ½ × 10⁻⁵ × 10⁴ = 0.05 J.
4. Q × 3, V same, E same (V and d unchanged), U × 3.
5. 8.85 × 10⁻¹² × 0.02/10⁻³ = 1.77 × 10⁻¹⁰ F = 177 pF.
6. C = ε₀A/(d − d/2) = 2C₀.
7. (a) Q = 300 + 240 = 540 μC, V = 540/9 = 60 V. (b) Q = 300 − 240 = 60 μC, V = 60/9 ≈ 6.7 V.
8. u = ½ × 8.85 × 10⁻¹² × 10¹² ≈ 4.4 J/m³.

What to do next

• Derive C = ε₀A/d and the slab formula ε₀A/(d − t + t/K) from the field between the plates.
• Make a one-page table of what stays fixed in each situation: battery on, battery off, charge sharing.
• Solve 15 network problems, including ones where you must spot a balanced bridge of capacitors.
• Move on to current electricity, and keep the physics formula sheet handy for revision.