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

Options

A. `2.5xx10^(-8)J`
B. `3.5xx10^(-6)J`
C. `2.5xx10^(-4)J`
D. `3.5xx10^(-4)J`

Answer

B

Explanation

Here `C = 700 pF = 700 xx 10^(-12) F,V = 100V ` Energy stored
` U = (1)/(2) CV^(2) = (1)/(2) xx 700 xx 10^(-12) xx (100)^(2) 3.5 xx 10^(-6) J`