Energy stored in capacitor 4μF charged to 100V

medium CBSE JEE-MAIN NEET • 2 min read
Tags Capacitors

Question

Calculate the energy stored in a capacitor of capacitance 4 μF4\ \mu\text{F} when it is charged to a potential difference of 100 V.

Solution — Step by Step

The energy stored in a capacitor is:

U=12CV2U = \frac{1}{2}CV^2

This can also be written as U=Q22C=QV2U = \frac{Q^2}{2C} = \frac{QV}{2}, but the 12CV2\frac{1}{2}CV^2 form is most convenient when CC and VV are given.

C=4 μF=4×10−6 FC = 4\ \mu\text{F} = 4 \times 10^{-6}\ \text{F}

V=100 VV = 100\ \text{V}

 U=12×4×10−6×(100)2U = \frac{1}{2} \times 4 \times 10^{-6} \times (100)^2 U=12×4×10−6×104U = \frac{1}{2} \times 4 \times 10^{-6} \times 10^4 U=12×4×10−2U = \frac{1}{2} \times 4 \times 10^{-2} U=2×10−2 JU = 2 \times 10^{-2}\ \text{J} U=0.02 J=20 mJ\boxed{U = 0.02\ \text{J} = 20\ \text{mJ}}

Why This Works

When a capacitor charges from 0 to voltage VV, the voltage doesn’t stay constant — it builds up gradually. The energy stored is the integral of power over time, which gives 12CV2\frac{1}{2}CV^2 rather than CV2CV^2. The factor of 12\frac{1}{2} accounts for this gradual buildup.

Physically, this energy is stored in the electric field between the plates. The energy density of the electric field is 12ϵ0E2\frac{1}{2}\epsilon_0 E^2 per unit volume — integrating this over the volume between the plates gives exactly 12CV2\frac{1}{2}CV^2.

Alternative Method — Using Charge

First find the charge: Q=CV=4×10−6×100=4×10−4 CQ = CV = 4 \times 10^{-6} \times 100 = 4 \times 10^{-4}\ \text{C}

Then: U=QV2=4×10−4×1002=4×10−22=0.02 JU = \frac{QV}{2} = \frac{4 \times 10^{-4} \times 100}{2} = \frac{4 \times 10^{-2}}{2} = 0.02\ \text{J} ✓

For JEE, remember all three equivalent forms: U=12CV2=Q22C=QV2U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{QV}{2}. Use whichever form matches the given data — if you’re given QQ and CC but not VV, use Q22C\frac{Q^2}{2C} directly to avoid an extra step.

Common Mistake

The most common error is forgetting the 12\frac{1}{2} factor and writing U=CV2U = CV^2. This gives double the correct answer. The factor of 12\frac{1}{2} is not optional — it comes directly from integrating dU=V dQdU = V\,dQ from 0 to QQ, and the average voltage during charging is V2\frac{V}{2}, not VV.

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