Two identical capacitors of 12 pF each are connected in series across a 50 V battery. Calculate the electrosta

Question

Q. Two identical capacitors of 12 pF each are connected in series across a 50 V battery. Calculate the electrostatic energy stored in the combination. If these were connected in parallel across the same battery, find out the value of the energy stored in this combination.

CBSECBSE Class XII Board 2019Subjective· 2mImportance★★★★★

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Concept understanding — Capacitor Energy Storage

Capacitor Energy Storage

The Intuition

Charging a capacitor is like piling sand onto a growing heap. The first grain of charge lands on an empty plate easily. But every later bit of positive charge must be pushed onto a plate that is already positive, and it resists. So more and more work is needed as the plate fills up. All of that work does not disappear — it is stored in the capacitor as electrostatic potential energy, ready to be released later.

Building the Formula

Suppose at some instant during charging the capacitor already holds charge $q$ , so the voltage across it at that moment is $v = q/C$ . Moving one more small charge $dq$ onto the plate costs work:

dW = v\,dq = \frac{q}{C}\,dq

Adding up (integrating) all these small contributions as the charge builds from $0$ to a final value $Q$ gives the total work done:

W = \int_0^{Q} \frac{q}{C}\,dq = \frac{1}{C}\cdot\frac{Q^2}{2} = \frac{Q^2}{2C}

This work is exactly the energy $U$ stored in the charged capacitor.

Three Equivalent Forms

Using $Q = CV$ , the same stored energy can be written three ways — pick whichever matches the quantities you know:

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

Use $\dfrac{Q^2}{2C}$ when the charge is fixed (capacitor disconnected from the source).
Use $\dfrac{1}{2}CV^2$ when the voltage is fixed (capacitor stays connected to a battery).

Watch out

The factor of $\tfrac{1}{2}$ is essential. A common error is writing $U = QV$ . That would only be true if the full voltage $V$ acted while all the charge moved — but the voltage climbs steadily from $0$ to $V$ as the plates fill, so the effective average voltage is $V/2$ , giving $U = \tfrac{1}{2}QV$ .

Where the Energy Lives — Energy Density

The energy is stored in the electric field occupying the space between the plates, not on the plates themselves. For a parallel-plate capacitor this leads to a general result: energy stored per unit volume of field is

u = \frac{1}{2}\,\varepsilon_0 E^2

where $E$ is the field strength. Wherever an electric field exists, energy is stored there, with density proportional to $E^2$ .

A Quick Example

A $10\ \mu\text{F}$ capacitor is charged to $100\ \text{V}$ . The stored energy is:

U = \frac{1}{2}CV^2 = \frac{1}{2}\times(10\times10^{-6})\times(100)^2 = 0.05\ \text{J}

That 0.05 J can be released almost instantly — which is exactly how a camera flash works: charge slowly, discharge fast. …

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For capacitors in series, the equivalent capacitance is half of each (6 pF), giving stored energy $U_s = \frac{1}{2} C_s V^2 = 7.5 \times 10^{-9} \text{ J}$ . For parallel, the equivalent capacitance doubles (24 pF), giving $U_p = \frac{1}{2} C_p V^2 = 3.0 \times 10^{-8} \text{ J}$ .

The core idea here is that the energy stored in a capacitor (or a combination) depends on two things: the equivalent capacitance of the network and the voltage across it. The formula $U = \frac{1}{2} C V^2$ is your starting point. But you must first find the right $C$ for the series and parallel cases.

A common mistake is to plug the individual capacitance values directly into the energy formula without first combining them. That would give the energy stored in one capacitor, not the whole network. The battery sees the combination as a single equivalent capacitor.

Let’s work through it.

1. Series combination

When two identical capacitors are connected in series, the equivalent capacitance is given by:

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

Since $C_1 = C_2 = 12 \text{ pF}$ :

\frac{1}{C_s} = \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6}

So $C_s = 6 \text{ pF}$ .

Watch out

For $n$ identical capacitors in series, the equivalent capacitance is $C/n$ , not $nC$ . This is a classic inversion trap.

The battery voltage $V = 50 \text{ V}$ appears across this series combination. The energy stored is:

U_s = \frac{1}{2} C_s V^2 = \frac{1}{2} \times (6 \times 10^{-12}) \times (50)^2

Compute step by step:

U_s = \frac{1}{2} \times 6 \times 10^{-12} \times 2500 = 3 \times 10^{-12} \times 2500 = 7500 \times 10^{-12} = 7.5 \times 10^{-9} \text{ J}

So $U_s = 7.5 \text{ nJ}$ .

Tip

Notice that in series, each capacitor gets half the voltage (25 V each), so the energy in each is $\frac{1}{2} \times 12 \times 10^{-12} \times (25)^2 = 3.75 \text{ nJ}$ . Two of them sum to 7.5 nJ — consistent.

2. Parallel combination

For two identical capacitors in parallel, the equivalent capacitance is simply the sum:

$C_p = C_1 + C_2 = 12 + 12 = 24 \text{ pF}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set SEM31 markMCQ
$Q. The capacitances of two conductors X and Y are 3C and C respectively. The conductor X is given a charge Q which the conductor shares with Y. The ratio of total shared energy to its total initial energy is$
(a) 9 : 16
(b) √3 : 2
(c) 3 : 4
(d) 4 : 3
›Reveal solutionSolution

Charge Q on X (capacitance 3C) redistributes over X and Y (total 4C in parallel). Energy = Q²/2C_total, so ratio = (1/4C)/(1/3C) = 3/4 — some energy is lost. Option (c).

Step 1 — initial energy (all charge Q on X, capacitance 3C):

U_i = Q²/(2·3C) = Q²/6C.

Step 2 — after connection, X and Y share charge at a common potential; total capacitance = 3C + C = 4C, and total charge is conserved at Q.

U_f = Q²/(2·4C) = Q²/8C.

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The capacity of a condenser is 2\times10⁻⁶ F and its potential is 200 V. The energy released on discharging it fully will be$
(a) 0.02 J
(b) 0.04 J
(c) 0.08 J
(d) 0.16 J.
›Reveal solutionSolution

Energy stored in a charged capacitor is $E = (1/2)CV^2$ ; substituting the given values gives 0.04 J.

When a capacitor of capacitance $C$ is charged to potential difference $V$ , it stores energy

$E = (1/2)CV^2$

This energy is released (as heat/spark) when the capacitor is fully discharged.

Given: $C = 2 \times 10^{-6}$ F, $V = 200$ V.

…

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CBSE 2019Set ANNUAL1 markMCQ
$Q. The stored energy of a capacitor charged to 100 V is 1 J. Capacitance of the capacitor is -$
(a) 2 × 10^4 F
(b) 2 × 10^-4 F
(c) 2 × 10^2 F
(d) 2 × 10^-2 F
›Reveal solutionSolution

C = 2U/V² = 2(1)/(100)² = 2 × 10⁻⁴ F.

The energy stored in a capacitor is

$U = \frac{1}{2}CV^2.$

Rearranging for C:

…

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Q.Two identical capacitors of 12 pF each are connected in series across a 50 V battery. Calculate the electrostatic energy stored in the combination. If these were connected in parallel across the same battery, find out the value of the energy stored in this combination.

1. Series combination

2. Parallel combination

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