A 12 ;pF capacitor is connected to a 50 ;V ba | vedclass

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Question

(D) Given capacitance,$C = 12 \;pF = 12 	imes 10^{-12} \;F$.Potential difference,$V = 50 \;V$.The electrostatic energy $E$ stored in a capacitor is g

Vedclass · Accepted Answer

$1.5 	imes 10^{-8} \;J$

Vedclass · Answer

(D) Given capacitance,$C = 12 \;pF = 12 	imes 10^{-12} \;F$.Potential difference,$V = 50 \;V$.The electrostatic energy $E$ stored in a capacitor is given by the formula:$E = \frac{1}{2} C V^2$Substituting the values:$E = \frac{1}{2} 	imes (12 	imes 10^{-12} \;F) 	imes (50 \;V)^2$$E = 6 	imes 10^{-12} 	imes 2500 \;J$$E = 15000 	imes 10^{-12} \;J$$E = 1.5 	imes 10^{-8} \;J$Therefore,the electrostatic energy stored in the capacitor is $1.5 	imes 10^{-8} \;J$.

$A$ $12 \;pF$ capacitor is connected to a $50 \;V$ battery. How much electrostatic energy is stored in the capacitor?

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