Capacitors

JEE Main level, single correct, medium

145 of 180

Question 145 of 180

The work done in placing a charge of 8×10−188 \times 10^{-18} coulomb on a condenser of capacity 100 micro-farad is

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Answer: (D) 32×10−3232 \times 10^{-32} joule

Work done =12q2c=(8×10−18)22×100×10−8=32×10−32 J=\frac{1}{2} \frac{\mathrm{q}^2}{\mathrm{c}}=\frac{\left(8 \times 10^{-18}\right)^2}{2 \times 100 \times 10^{-8}}=32 \times 10^{-32} \mathrm{~J}

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