Capacitors and Capacitance
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Direct answer
Capacitance is charge stored per unit potential difference, C = Q/V; for a parallel-plate capacitor C = epsilon_0 A/d, multiplied by K when a dielectric of constant K fills the gap. Stored energy is U = (1/2) C V^2 = Q^2/(2C) = (1/2) Q V. Series capacitors share charge and combine as 1/C = 1/C1 + 1/C2; parallel capacitors share voltage and add directly, C = C1 + C2.
What you must remember
- Key results: parallel-plate C = epsilon_0 A/d; isolated sphere C = 4 pi epsilon_0 R; a dielectric slab of constant K multiplies capacitance by K.
- Energy: U = (1/2) C V^2 = Q^2/(2C) = (1/2) Q V; energy density in any electric field = (1/2) epsilon_0 E^2.
- Series: same charge on each, voltages divide inversely with capacitance, 1/C_eq = 1/C1 + 1/C2 + ...; parallel: same voltage, charges divide in proportion, C_eq = C1 + C2 + ...
- Two charged capacitors joined: common voltage = (C1 V1 + C2 V2)/(C1 + C2); energy lost = C1 C2 (V1 − V2)^2/(2 (C1 + C2)) — dissipated in the wires and radiated, zero only when V1 = V2.
- Isolated capacitor (Q fixed): inserting a dielectric raises C and lowers V, E and stored energy — the slab is pulled in. Battery connected (V fixed): Q and stored energy both rise, the battery supplying the difference.
- Attractive force between plates: F = Q^2/(2 epsilon_0 A) — each plate sits in half the total field, that of the other plate alone.
- n identical plates alternately connected give (n − 1) times the capacitance of a single pair.
Common confusion
The recurring error is never deciding what stays fixed. Disconnected from the battery, Q is constant and every change tracks through Q^2/(2C); still connected, V is constant and changes track through (1/2) C V^2 with the battery moving charge. Students who blur the two get the direction of energy change backwards. The energy loss on joining capacitors also seems paradoxical — the missing energy leaves as heat and electromagnetic radiation, whatever the connecting resistance.
Exam-focused takeaway
JEE Main tests equivalent capacitance of networks, stored energy, charge division and the dielectric multiplier — dependable marks when series/parallel identification is secure. JEE Advanced prefers the conceptual edge: a slab partially inserted (fringing fields pull it in), a switch reconnecting charged capacitors (charge conservation on isolated plate islands), capacitor bridges collapsed by symmetry, and the battery's work versus the change in stored energy during insertion. Ask "what is held constant?" before writing any energy expression.
Frequently asked questions
Why is energy lost when two charged capacitors are connected?
Charge redistributes and the stored energy falls by C1 C2 (V1 − V2)^2/(2 (C1 + C2)), dissipated as heat and radiation; the resistance sets only the rate, not the loss.
What stays constant when a dielectric is inserted with the battery connected?
The voltage — the battery fixes it; capacitance and charge increase, and the battery supplies the extra charge and energy.
And if the capacitor is isolated first?
The charge stays fixed; capacitance rises while voltage, field and stored energy all fall, and the slab is drawn into the gap.
Do series capacitors carry the same charge?
Yes — the plates between neighbours form isolated conductors that force equal and opposite induced charges, so the same charge passes through each while voltages divide.
Why does the force between plates use half the field?
Each plate sits in the field of the other plate alone, Q/(2 epsilon_0 A) — half the total field between the plates; using the full field double-counts and is the standard slip.