A 12,pF capacitor is connected to a 50,V batt | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The electrostatic energy $U$ stored in a capacitor is given by the formula: $U = \frac{1}{2}CV^2$.Given:Capacitance $C = 12\,pF = 12 	imes 10^{-12}\,F$.Potential difference $V = 50\,V$.Substituting the values into the formula:$U = \frac{1}{2} 	imes (12 	imes 10^{-12}) 	imes (50)^2$$U = 6 	imes 10^{-12} 	imes 2500$$U = 6 	imes 2500 	imes 10^{-12}$$U = 15000 	imes 10^{-12}\,J$$U = 1.5 	imes 10^4 	imes 10^{-12}\,J$$U = 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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