A 10 , pF capacitor is connected to a 50 , V | vedclass

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Question

(A) The energy $U$ stored in a capacitor is given by the formula $U = \frac{1}{2} C V^2$.Given:Capacitance $C = 10 \, pF = 10 	imes 10^{-12} \, F = 1

Vedclass · Accepted Answer

$1.25 	imes 10^{-8} \, J$

Vedclass · Answer

(A) The energy $U$ stored in a capacitor is given by the formula $U = \frac{1}{2} C V^2$.Given:Capacitance $C = 10 \, pF = 10 	imes 10^{-12} \, F = 10^{-11} \, F$.Potential difference $V = 50 \, V$.Substituting the values into the formula:$U = \frac{1}{2} 	imes (10 	imes 10^{-12} \, F) 	imes (50 \, V)^2$$U = \frac{1}{2} 	imes 10^{-11} 	imes 2500$$U = 0.5 	imes 2.5 	imes 10^{-8} \, J$$U = 1.25 	imes 10^{-8} \, J$.Therefore,the correct option is $A$.

$A$ $10 \, pF$ capacitor is connected to a $50 \, V$ battery. How much electrostatic energy is stored in the capacitor?

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