A capacitor with capacitance $5 \mu \mathrm{~F}$ is charged to $5 \mu \mathrm{C}$. If the plates are pulled apart to reduce the capacitance to $2 \mu \mathrm{~F}$, how much work is done?

(A) $3.75 \times 10^{-6} \mathrm{~J}$

(B) $2.55 \times 10^{-6} \mathrm{~J}$

(C) $6.25 \times 10^{-6} \mathrm{~J}$

(D) $2.16 \times 10^{-6} \mathrm{~J}$

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Capacitance

A capacitor with capacitance

5 μ F

is charged to

5 μ C

. If the plates are pulled apart to reduce the capacitance to

2 μ F

, how much work is done?

(A) $3.75 \times 1 0^{- 6} J$ ✓ Correct
(B) $2.55 \times 1 0^{- 6} J$
(C) $6.25 \times 1 0^{- 6} J$
(D) $2.16 \times 1 0^{- 6} J$

Answer: (A)

Solution

Pulling the plates apart leaves the charge fixed, so the work done equals the increase in stored energy

\frac{q ^{2}}{2 C}

. With

q = 5 μ C

W = \frac{q ^{2}}{2} (\frac{1}{C _{f}} - \frac{1}{C _{i}}) = \frac{( 5 \times 1 0 ^{- 6} ) ^{2}}{2} (\frac{1}{2 \times 1 0 ^{- 6}} - \frac{1}{5 \times 1 0 ^{- 6}})

. The bracket is

\frac{3}{10 \times 1 0 ^{- 6}}

, so

W = \frac{25 \times 1 0 ^{- 12}}{2} \times \frac{3}{10 \times 1 0 ^{- 6}} = 3.75 \times 1 0^{- 6}

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option

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