Capacitors — Storing and Releasing Charge

A capacitor is the simplest device that stores electrical energy — not as current flowing, but as charge sitting on two conducting plates. Understanding why charge accumulates, what limits it, and how capacitors combine is the foundation for understanding filters, amplifiers, power supplies, and every electronic circuit.

The physics is elegant: two conductors separated by an insulator. Charge one plate positive, the other goes negative, and you’ve stored energy in the electric field between them.

Key Terms & Definitions

Capacitance ( $C$ ): The ratio of charge stored to voltage across the capacitor. $C = Q/V$ . Measured in Farads (F). A 1 F capacitor holds 1 Coulomb of charge at 1 Volt — enormous in practice; most capacitors are in μF, nF, or pF range.

Dielectric: An insulating material placed between capacitor plates. It increases capacitance by reducing the electric field between plates (the dielectric becomes polarised and partially cancels the field). Common dielectrics: air, mica, ceramic, paper.

Dielectric constant (κ or K): The factor by which capacitance increases when a dielectric fills the space between plates. For vacuum/air, κ = 1. For mica, κ ≈ 5–8. For water, κ ≈ 80.

Capacitance of a parallel plate capacitor:

C = \frac{\kappa \epsilon_0 A}{d}

where $A$ is plate area, $d$ is plate separation, $\epsilon_0 = 8.85 \times 10^{-12}\text{ F/m}$ is permittivity of free space.

Energy stored in a capacitor:

U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{QV}{2}

All three forms are equivalent — choose based on what quantities you’re given.

How Capacitors Work — The Intuition

When you connect a capacitor to a battery, electrons flow from one plate to the other through the external circuit (not through the gap). One plate accumulates excess electrons (–Q), the other loses electrons (+Q). This creates an electric field between the plates.

As charge builds up, the electric field grows stronger and opposes further charge flow — eventually the voltage across the capacitor equals the battery voltage, and charge flow stops. The capacitor is now “charged.”

The stored energy sits in this electric field. When the battery is disconnected and a resistor is connected, the field drives charge back around the circuit — the capacitor discharges and delivers its stored energy.

Series and Parallel Combinations

\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}

Same charge on each capacitor; voltages add.

$C_{eq}$ is always less than the smallest individual capacitor.

C_p = C_1 + C_2 + C_3

Same voltage across each capacitor; charges add.

$C_{eq}$ is always greater than the largest individual capacitor.

Why the rules are “opposite” to resistors: Capacitance is like “storage space.” Plates in parallel add more storage area — so capacitance adds. Plates in series share charge but divide voltage — the total storage is limited by the smallest, so reciprocals add.

Effect of a Dielectric

When a dielectric slab is inserted between fully charged plates (with battery disconnected vs. connected — two very different scenarios):

Battery disconnected (charge Q is constant):

Q stays the same (no path for charge to move)
Capacitance increases: $C' = \kappa C$
Voltage decreases: $V' = V/\kappa$
Energy decreases: $U' = U/\kappa$ (energy is extracted by the work done pulling in the dielectric)

Battery connected (voltage V is constant):

V stays the same (battery maintains it)
Capacitance increases: $C' = \kappa C$
Charge increases: $Q' = \kappa Q$
Energy increases: $U' = \kappa U$ (extra energy comes from the battery)

JEE Main has a classic problem where a dielectric is inserted and you must determine whether the battery is connected or not before calculating how Q, V, and U change. Always identify the constant quantity first — if battery is connected, V is constant; if disconnected, Q is constant.

Solved Examples

Example 1 — CBSE Level: Series combination

Q: Three capacitors of 2 μF, 3 μF, and 6 μF are connected in series. Find the equivalent capacitance and the charge on each capacitor when 12 V is applied.

Solution:

\frac{1}{C_s} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = 1 \Rightarrow C_s = 1\,\mu\text{F}

In series, same charge on each: $Q = C_s V = 1 \times 12 = 12\,\mu\text{C}$

Voltages: $V_1 = 12/2 = 6\text{ V}$ ; $V_2 = 12/3 = 4\text{ V}$ ; $V_3 = 12/6 = 2\text{ V}$ . Check: $6+4+2 = 12$ V ✓

Example 2 — JEE Main Level: Energy stored

Q: A 10 μF capacitor is charged to 200 V. Find the energy stored. If it is connected in parallel with an uncharged 10 μF capacitor (with no battery), find the final energy and explain the difference.

Solution:

Initial: $U_i = \frac{1}{2}CV^2 = \frac{1}{2}(10 \times 10^{-6})(200)^2 = 0.2\text{ J}$ . Charge: $Q = 10 \times 10^{-6} \times 200 = 2\text{ mC}$ .

When connected in parallel: total capacitance = 20 μF. Charge redistributes: $V_f = Q/C_{total} = 2\text{ mC}/20\,\mu\text{F} = 100\text{ V}$ .

Final energy: $U_f = \frac{1}{2}(20 \times 10^{-6})(100)^2 = 0.1\text{ J}$ .

Energy lost = 0.2 – 0.1 = 0.1 J. This energy is dissipated as heat in the connecting wires (even “ideal” wires have some resistance, and the charge redistribution current generates heat).

Example 3 — JEE Advanced Level: Capacitor with dielectric slab

Q: A parallel plate capacitor ( $C = 4\,\mu\text{F}$ ) is connected to a 12 V battery. A dielectric slab of $\kappa = 3$ is then inserted fully between the plates. Find the new charge, voltage, and energy. Is energy gained or lost from the battery?

Solution:

Original: $Q_0 = 4 \times 12 = 48\,\mu\text{C}$ , $U_0 = \frac{1}{2}(4)(12)^2 = 288\,\mu\text{J}$

After inserting dielectric (battery remains connected, V constant): $C' = 3 \times 4 = 12\,\mu\text{F}$ $V' = 12\text{ V}$ (unchanged, battery connected) $Q' = 12 \times 12 = 144\,\mu\text{C}$ $U' = \frac{1}{2}(12)(12)^2 = 864\,\mu\text{J}$

Extra charge from battery: $\Delta Q = 144 - 48 = 96\,\mu\text{C}$ . Work done by battery: $W_{battery} = \Delta Q \times V = 96 \times 12 = 1152\,\mu\text{J}$ . Energy to capacitor: $\Delta U = 864 - 288 = 576\,\mu\text{J}$ . Energy to dielectric insertion work: $1152 - 576 = 576\,\mu\text{J}$ is done mechanically (the slab is pulled in by electrostatic force).

Exam-Specific Tips

CBSE Class 12 (2–5 mark questions): Derivation of capacitance of parallel plate capacitor, effect of dielectric, series/parallel equivalent. Expect at least 1 numerical on energy stored or charge distribution in combination circuits.

JEE Main: Numericals on energy conservation in capacitor-battery systems, redistribution of charge, dielectric insertion scenarios. The “battery connected vs. disconnected” distinction appears in almost every paper.

NEET: Conceptual MCQs on which quantity remains constant (Q or V) in different situations, and qualitative effect of dielectric insertion. Direct formula substitution questions.

Common Mistakes to Avoid

Mistake 1: Using the series formula as $C_s = C_1 + C_2 + C_3$ (the parallel formula). Remember: for capacitors in series, reciprocals add; for parallel, capacitances add. The opposite of the resistor rule.

Mistake 2: After charge redistribution, using the original voltage. When two capacitors are connected, the final voltage is $V_f = Q_{total}/C_{total}$ — always recalculate voltage, don’t assume it stays the same.

Mistake 3: Confusing “charge is constant” (battery disconnected) with “voltage is constant” (battery connected). This single confusion is responsible for half the errors in dielectric problems.

Mistake 4: Forgetting that energy stored in a capacitor is $\frac{1}{2}CV^2$ , not $CV^2$ . The factor of $\frac{1}{2}$ is because the charge builds up gradually from 0 to Q — the average voltage during charging is V/2.

Mistake 5: When capacitors are connected in parallel to a battery and the battery is then removed, students assume the total charge doubles. No — charge is conserved. The charge that was on the first capacitor redistributes, and no new charge enters the system once the battery is disconnected.

Practice Questions

Q1. A 4 μF capacitor is charged to 100 V. Calculate the energy stored and the charge.

$Q = CV = 4 \times 10^{-6} \times 100 = 400\,\mu\text{C}$ . $U = \frac{1}{2}CV^2 = \frac{1}{2}(4 \times 10^{-6})(100)^2 = 0.02\text{ J} = 20\text{ mJ}$ .

Q2. Two capacitors of 6 μF and 3 μF are connected in series across 90 V. Find voltage across each.

$C_s = \frac{6 \times 3}{6+3} = 2\,\mu\text{F}$ . Charge $Q = 2 \times 90 = 180\,\mu\text{C}$ (same for series). $V_1 = Q/C_1 = 180/6 = 30\text{ V}$ . $V_2 = Q/C_2 = 180/3 = 60\text{ V}$ . Check: $30+60=90$ ✓.

Q3. A parallel plate capacitor has plates of area 100 cm² separated by 2 mm. Find capacitance (dielectric constant = 1, $\epsilon_0 = 8.85 \times 10^{-12}$ F/m).

$C = \epsilon_0 A/d = (8.85 \times 10^{-12})(100 \times 10^{-4})/(2 \times 10^{-3}) = 4.425 \times 10^{-12}\text{ F} \approx 4.4\text{ pF}$ .

Q4. How does capacitance change if plate separation is doubled and plate area is also doubled?

$C = \epsilon_0 A/d$ . New capacitance $= \epsilon_0 (2A)/(2d) = \epsilon_0 A/d = C$ . Capacitance remains unchanged.

Q5. A capacitor of capacitance $C$ is charged to $V_0$ and disconnected. A second identical uncharged capacitor is connected in parallel. Find the final energy and explain why it is less than the initial energy.

Initial: $Q_0 = CV_0$ , $U_i = \frac{1}{2}CV_0^2$ . After: $V_f = Q_0/(2C) = V_0/2$ . $U_f = \frac{1}{2}(2C)(V_0/2)^2 = \frac{1}{4}CV_0^2 = \frac{1}{2}U_i$ . Half the energy is lost as heat during charge redistribution through the connecting wires.

FAQs

Why is the unit of capacitance called a Farad?

Named after Michael Faraday (1791–1867), who pioneered the study of electromagnetic induction and electrochemistry. 1 Farad is an enormous capacitance — most practical capacitors are measured in microfarads or picofarads.

Why does inserting a dielectric increase capacitance?

The dielectric molecules get polarised by the electric field and create their own opposing field. This reduces the net field between the plates. Since $V = Ed$ and Q is constant (disconnected), lower E means lower V. Since $C = Q/V$ , lower V means higher C.

Can a capacitor hold charge indefinitely?

In theory yes, but in practice no. Real capacitors have some leakage — the dielectric is not a perfect insulator and slowly conducts tiny currents. Electrolytic capacitors (the cylindrical ones in electronic circuits) discharge noticeably over months; ceramic capacitors hold charge much longer.

What’s the difference between a capacitor and a battery?

A battery stores energy in chemical form and releases it at roughly constant voltage over long periods. A capacitor stores energy in an electric field and can release it almost instantly (very high power delivery), but stores far less total energy than a battery of similar volume. Capacitors are used for rapid charge/discharge applications; batteries for sustained energy delivery.

Capacitor Networks — Beyond Simple Series/Parallel

Many JEE problems involve capacitor networks that are neither purely series nor purely parallel. The strategy is to simplify step by step, or use symmetry.

Wheatstone Bridge for Capacitors

When four capacitors are arranged in a Wheatstone bridge configuration with a fifth capacitor across the “bridge,” and the bridge is balanced ( $C_1/C_2 = C_3/C_4$ ), then no charge flows through the bridge capacitor and it can be removed from the circuit.

Worked Example — Mixed Network

Three capacitors: $C_1 = 2\,\mu\text{F}$ , $C_2 = 4\,\mu\text{F}$ , $C_3 = 6\,\mu\text{F}$ . $C_1$ and $C_2$ are in series, and their combination is in parallel with $C_3$ . Find the equivalent capacitance.

$\frac{1}{C_{12}} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4} \implies C_{12} = \frac{4}{3}\,\mu\text{F}$

$C_{eq} = C_{12} + C_3 = \frac{4}{3} + 6 = \frac{4 + 18}{3} = \frac{22}{3} \approx 7.33\,\mu\text{F}$

Energy Density of Electric Field

The energy stored in a capacitor is distributed throughout the electric field between the plates. The energy per unit volume (energy density) is:

u = \frac{1}{2}\epsilon_0 E^2 = \frac{1}{2}\epsilon_0 \kappa E^2 \text{ (with dielectric)}

where $E$ is the electric field strength.

For a parallel plate capacitor: $E = V/d$ , so $u = \frac{1}{2}\epsilon_0(V/d)^2$

Total energy $= u \times \text{volume} = \frac{1}{2}\epsilon_0 \frac{V^2}{d^2} \times Ad = \frac{1}{2}\frac{\epsilon_0 A}{d}V^2 = \frac{1}{2}CV^2$ — consistent with our earlier formula.

JEE Advanced 2023 had a problem on energy density of the electric field between capacitor plates. The formula $u = \frac{1}{2}\epsilon_0 E^2$ is fundamental — it applies not just to capacitors but to any region of space where an electric field exists. When a dielectric is present, $\epsilon_0$ is replaced by $\epsilon_0\kappa$ .

Q6. Two capacitors $C$ and $2C$ are charged to voltages $2V$ and $V$ respectively, then connected positive-to-positive. Find the final voltage and charge on each.

Initial charges: $Q_1 = C \times 2V = 2CV$ , $Q_2 = 2C \times V = 2CV$ .

Total charge (since positive plates are connected): $Q_{total} = 2CV + 2CV = 4CV$ .

Total capacitance: $C + 2C = 3C$ .

Final voltage: $V_f = \frac{4CV}{3C} = \frac{4V}{3}$ .

Charge on $C$ : $q_1 = C \times \frac{4V}{3} = \frac{4CV}{3}$ . Charge on $2C$ : $q_2 = 2C \times \frac{4V}{3} = \frac{8CV}{3}$ .

Check: $q_1 + q_2 = 4CV$ . Correct.

Q7. A capacitor of capacitance 5 μF is charged to 10 V. It is then connected to an uncharged capacitor of 20 μF. What fraction of the initial energy is lost?

$U_i = \frac{1}{2}(5)(100) = 250\,\mu\text{J}$ . $Q = 50\,\mu\text{C}$ .

After connecting: $V_f = 50/(5+20) = 2$ V. $U_f = \frac{1}{2}(25)(4) = 50\,\mu\text{J}$ .

Fraction lost $= \frac{250 - 50}{250} = \frac{200}{250} = \frac{4}{5} = 80\%$ . Four-fifths of the energy is dissipated as heat.