If the energy stored in a spherical conductor | vedclass

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(A) The energy $U$ stored in a spherical conductor of radius $R$ carrying a charge $Q$ is given by the formula: $U = \frac{Q^2}{2C}$.For a spherical c

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$10.8$

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(A) The energy $U$ stored in a spherical conductor of radius $R$ carrying a charge $Q$ is given by the formula: $U = \frac{Q^2}{2C}$.For a spherical conductor,the capacitance $C$ is given by $C = 4 \pi \epsilon_0 R$.Substituting $C$ into the energy formula: $U = \frac{Q^2}{2(4 \pi \epsilon_0 R)} = \frac{Q^2}{8 \pi \epsilon_0 R}$.Given $Q = 12 	imes 10^{-6} \ C$ and $U = 6 \ J$.We know $\frac{1}{4 \pi \epsilon_0} = 9 	imes 10^9 \ N \cdot m^2/C^2$.Rearranging for $R$: $R = \frac{Q^2}{8 \pi \epsilon_0 U} = \frac{Q^2}{2U} 	imes \frac{1}{4 \pi \epsilon_0}$.$R = \frac{(12 	imes 10^{-6})^2}{2 	imes 6} 	imes 9 	imes 10^9$.$R = \frac{144 	imes 10^{-12}}{12} 	imes 9 	imes 10^9 = 12 	imes 10^{-12} 	imes 9 	imes 10^9$.$R = 108 	imes 10^{-3} \ m = 0.108 \ m = 10.8 \ cm$.

If the energy stored in a spherical conductor having a charge of $12 \mu C$ is $6 \ J$,then the radius of the spherical conductor is (in $cm$)

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