A capacitor has a capacity of 2,mu F and is c | vedclass

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(D) The energy stored in a capacitor is given by the formula $U = \frac{1}{2}CV^2$.Given: $C = 2\,\mu F = 2 	imes 10^{-6}\,F$ and $V = 50\,V$.Substit

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$25 	imes 10^3\,erg$

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(D) The energy stored in a capacitor is given by the formula $U = \frac{1}{2}CV^2$.Given: $C = 2\,\mu F = 2 	imes 10^{-6}\,F$ and $V = 50\,V$.Substituting the values: $U = \frac{1}{2} 	imes (2 	imes 10^{-6}) 	imes (50)^2$.$U = 10^{-6} 	imes 2500 = 2500 	imes 10^{-6} = 2.5 	imes 10^{-3}\,J$.Since $1\,J = 10^7\,erg$,we convert the energy to $erg$:$U = 2.5 	imes 10^{-3} 	imes 10^7\,erg = 2.5 	imes 10^4\,erg$.Note: Based on the provided options,the intended calculation likely assumes $1\,J = 10^6\,erg$ or a specific unit conversion context. Recalculating with the provided option $25 	imes 10^3\,erg$ as the target: $2.5 	imes 10^{-3}\,J = 25 	imes 10^{-4}\,J$. Converting to $erg$ using $1\,J = 10^7\,erg$ gives $25 	imes 10^3\,erg$.

$A$ capacitor has a capacity of $2\,\mu F$ and is charged to a voltage of $50\,V$. The energy stored is:

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