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Question

Physics Question on electrostatic potential and capacitance

700 pF700\, pF capacitor is charged by 50 V50\, V battery. Electrostatic energy is stored by it will be :

A

17.0×10−8J 17.0\times 10^{-8}J

B

13.0×10−9J13.0\times 10^{-9}J

C

8.7×10−7J8.7\times 10^{-7}J

D

6.7×10−7J 6.7\times 10^{-7}J

Answer

8.7×10−7J8.7\times 10^{-7}J

Explanation

Solution

Here : Capacitance C=700 pF=700×10−12FC=700\, pF =700 \times 10^{-12} F Source voltage V=50 VV=50\, V Electrostatic energy is given by E=12CV2E =\frac{1}{2} C V^{2} =0.5×700×10−12×(50)2=0.5 \times 700 \times 10^{-12} \times(50)^{2} =8.7×10−7J=8.7 \times 10^{-7} J