The capacity of a parallel plate air capacitor is `10muF` and it is given a charge of `40muC`. The electrical energy stored in the capacitor is
A. 400 erg
B. 600 erg
C. 800 erg
D. 900 erg

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1 Answers

Correct Answer - C
`E=Q^(2)/(2C)=((40xx10^(-6))^(2))/(2xx10xx10^(-6))`
`=80xx10^(-6)J`
= 800 erg.

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