A 12 text{ pF} capacitor is connected to a 50 | vedclass

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Question

(A) Given:Capacitance $C = 12 	ext{ pF} = 12 	imes 10^{-12} 	ext{ F}$Potential difference $V = 50 	ext{ V}$The electrostatic energy $U$ stored in

Vedclass · Accepted Answer

$1.5 	imes 10^{-8}$

Vedclass · Answer

(A) Given:Capacitance $C = 12 	ext{ pF} = 12 	imes 10^{-12} 	ext{ F}$Potential difference $V = 50 	ext{ V}$The electrostatic energy $U$ stored in a capacitor is given by the formula:$U = \frac{1}{2} C V^2$Substituting the given values:$U = \frac{1}{2} 	imes (12 	imes 10^{-12} 	ext{ F}) 	imes (50 	ext{ V})^2$$U = 6 	imes 10^{-12} 	imes 2500$$U = 15000 	imes 10^{-12} 	ext{ J}$$U = 1.5 	imes 10^4 	imes 10^{-12} 	ext{ J}$$U = 1.5 	imes 10^{-8} 	ext{ J}$Therefore,the energy stored is $1.5 	imes 10^{-8} 	ext{ J}$.

$A$ $12 \text{ pF}$ capacitor is connected to a $50 \text{ V}$ battery. The electrostatic energy stored in the capacitor will be . . . . . . $\text{J}$.

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