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2. Electric Potential and Capacitance
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A piece of cloud having area $25 \times {10^6}\,{m^2}$ and electric potential of ${10^5}$ $volts$. If the height of cloud is $0.75\,km$, then energy of electric field between earth and cloud will be.....$J$
A
$250$
B
$750$
C
$1225$
D
$1475$
Solution
(d)Energy $ = \frac{1}{2}{\varepsilon _0}{E^2} \times (A \times d)$$ = \frac{1}{2}{\varepsilon _0}\left( {\frac{{{V^2}}}{{{d^2}}}} \right)\,Ad$
$ = \frac{1}{2} \times \frac{{8.85 \times {{10}^{ – 12}} \times {{({{10}^5})}^2} \times 25 \times {{10}^6}}}{{0.75 \times {{10}^3}}} = 1475\,J$
Standard 12
Physics
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