Question

A capacitor of capacitor of capacitance 700 pF is charged by 100 V battery. The electrostatic energy stored by the capacitor is:

• A. $2.5\times {10}^{-8}J$
• B. $3.5\times {10}^{-6}J$
• C. $2.5\times {10}^{-4}J$
• D. $3.5\times {10}^{-4}J$

Answer 1

Answer: (B)

$3.5\times {10}^{-6}J$

Here, $C=700pF=700\times {10}^{-12}F,V=100V$
$U=\frac{1}{2}C{V}^2=\frac{1}{2}\times 700\times {10}^{-12}\times {\left(100\right)}^2=3.5\times {10}^{-6}J$