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Question

Physics Question on electrostatic potential and capacitance

A 16 pF16 \,pF capacitor is connected to 70 V70\, V supply. The amount of electric energy stored in the capacitor is

A

4.5×10−12 J4.5 \times 10^{-12} \, J

B

5.1×10−8 J5.1 \times 10^{-8} \, J

C

2.5×10−12 J2.5 \times 10^{-12} \, J

D

3.9×10−8 J3.9 \times 10^{-8} \, J

Answer

3.9×10−8 J3.9 \times 10^{-8} \, J

Explanation

Solution

Energy stored =12CV2=\frac{1}{2} C V^{2} =12(16×10−12)×70×70=\frac{1}{2}\left(16 \times 10^{-12}\right) \times 70 \times 70 =392×10−10=392 \times 10^{-10} =3.92×10−8J=3.92 \times 10^{-8} J