Question

The work done in placing a charge of $8\times {10}^{-18}$ coulomb on a condenser of capacity 100 micro-Farad is:
[AIEEE-2003]

• A. $16\times {10}^{-32}joule$
• B. $3.1\times {10}^{-26}joule$
• C. $4\times {10}^{-10}joule$
• D. $32\times {10}^{-32}joule$

Answer 1

The correct option is D $32\times {10}^{-32}joule$

Given, $Q=8\times {10}^{-18}C$
$C=100\times {10}^{-6}F$
$={10}^{-4}F$

work done = $\frac{Q^2}{2C}=\frac{64\times {10}^{-36}}{2\times {10}^{-4}}$
$=32\times {10}^{-32}$ Joules.