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Question

Physics Question on electrostatic potential and capacitance

A capacitor of capacitance 700 pF700\, pF is charged by 100 V100\, V battery. The electrostatic energy stored by the capacitor is

A

2.5×10−8 J2.5 \times 10^{-8} \, J

B

3.5×10−6 J3.5 \times 10^{-6} \, J

C

2.5×10−4 J2.5 \times 10^{-4} \, J

D

3.5×10−4 J3.5 \times 10^{-4} \, J

Answer

3.5×10−6 J3.5 \times 10^{-6} \, J

Explanation

Solution

Here, C=700 PF=700×10−12 FC=700 \, PF =700\times10^{-12}\, F, V=100 VV=100\,V Energy stored U=12CV2=12×700×10−12×(100)2U=\frac{1}{2}CV^{2}=\frac{1}{2}\times700\times10^{-12}\times\left(100\right)^{2} =3.5×10−6 J=3.5\times10^{-6}\, J