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2. Electric Potential and Capacitance
easy
A $700\,pF$ capacitor is charged by a $50\,V$ battery. The electrostatic energy stored by it is
A
$17.0 \times {10^{ - 8}}\,J$
B
$13.6 \times {10^{ - 9}}\,J$
C
$9.5 \times {10^{ - 9}}\,J$
D
$8.7 \times {10^{ - 7}}\,J$
Solution
(d) $U = \frac{1}{2}C{V^2} = \frac{1}{2} \times 700 \times {10^{ – 12}}{(50)^2}$$ = 8.7 \times {10^{ – 7}}\,J$
Standard 12
Physics
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