The work done in placing a charge of $8 \times 10^{-18}$ coulomb on a condenser of capacity 100 micro-farad is
The work done in placing a charge of $8 \times 10^{-18}$ coulomb on a condenser of capacity 100 micro-farad is
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$16 \times 10^{-32}$ joule
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$3.1 \times 10^{-26}$ joule
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$4 \times 10^{-10}$ joule
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$32 \times 10^{-32}$ joule
Solution
Work done $=\frac{1}{2} \frac{\mathrm{q}^2}{\mathrm{c}}=\frac{\left(8 \times 10^{-18}\right)^2}{2 \times 100 \times 10^{-8}}=32 \times 10^{-32} \mathrm{~J}$
Asked in: JEE Main 2003
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