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

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

A

2.5×10−8J2.5 \times 10^{- 8}J

B

3.5×10−6J3.5 \times 10^{- 6}J

C

2.5×10−4J2.5 \times 10^{- 4}J

D

3.5×10−4J3.5 \times 10^{- 4}J

Answer

3.5×10−6J3.5 \times 10^{- 6}J

Explanation

Solution

: Here, C=700pF=700×10−12F,V=100VC = 700pF = 700 \times 10^{- 12}F,V = 100V

Energy stored

U=12CV2=12×700×10−12×(100)2=3.5×10−6JU = \frac{1}{2}CV^{2} = \frac{1}{2} \times 700 \times 10^{- 12} \times (100)^{2} = 3.5 \times 10^{- 6}J