JEE MAIN - Physics (2003 - No. 51)
The work done in placing a charge of $$8 \times {10^{ - 18}}$$ coulomb on a condenser of capacity $$100$$ micro-farad is
$$16 \times {10^{ - 32}}\,\,joule$$
$$3.1 \times {10^{ - 26}}\,\,joule$$
$$4 \times {10^{ - 10}}\,\,joule$$
$$32 \times {10^{ - 32}}\,\,joule$$
Explanation
The work done is stored as the potential energy. The potential energy stored in a capacitor is given by
$$U = {1 \over 2}{{{Q^2}} \over C}$$
$$ = {1 \over 2} \times {{{{\left( {8 \times {{10}^{ - 18}}} \right)}^2}} \over {100 \times {{10}^{ - 6}}}}$$
$$ = 32 \times {10^{ - 32}}J$$
$$U = {1 \over 2}{{{Q^2}} \over C}$$
$$ = {1 \over 2} \times {{{{\left( {8 \times {{10}^{ - 18}}} \right)}^2}} \over {100 \times {{10}^{ - 6}}}}$$
$$ = 32 \times {10^{ - 32}}J$$
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