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Question
A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor?
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Solution
Given: Capacitor of the capacitance, C = 12 pF = 12 × 10−12 F
Potential difference, V = 50 V
Formula: Electrostatic energy stored in the capacitor is given by the relation,
`E = 1/2 CV^2`
= `1/2 xx 12 xx 10^-12 xx (50)^2`
= `1/2 xx 12 xx 10^-12 xx 2500`
= `(30000 xx 10^-12)/2`
= 15000 × 10−12
= 1.5 × 10−8 J
Therefore, the electrostatic energy stored in the capacitor is 1.5 × 10−8 J.
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