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2. Electric Potential and Capacitance
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A charge of $40\,\mu \,C$ is given to a capacitor having capacitance $C = 10\,\mu \,F$. The stored energy in ergs is
A
$80 \times {10^{ - 6}}$
B
$800$
C
$80$
D
$8000$
Solution
(b) $U = \frac{{{Q^2}}}{{2C}} = \frac{{{{(40 \times {{10}^{ – 6}})}^2}}}{{2 \times {{10}^{ – 6}} \times 10}} = \frac{{16 \times {{10}^{ – 10}}}}{{2 \times {{10}^{ – 5}}}} = 8 \times {10^{ – 5}}\,J$
$ = 8 \times {10^{ – 5}} \times {10^7} = 800\,erg$
Standard 12
Physics
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