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Solution :

Capacitor of the capacitance , C=12 pF =`12xx10^(-12)F` <br> potential difference, V=50 V <br> Electrostatic energy stored in the capacitor is given by the relation, <br> `E=1/2 CV^(2)` <br> `=1/2 xx12 xx10^(-12) xx(50)^(2)` <br> `=1.5xx10^(-8) J` <br> Therefore, the electrostatic energy stored in the capacitor is `1.5xx10^(-8)J`.Related Video

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