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(B) The energy stored in a capacitor is given by $U = \frac{1}{2}CV^2$.Given: Capacitance $C = 6 	imes 10^{-6} \,F$,initial potential $V_1 = 10 \,V$,

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$9 	imes 10^{-4} \,J$

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(B) The energy stored in a capacitor is given by $U = \frac{1}{2}CV^2$.Given: Capacitance $C = 6 	imes 10^{-6} \,F$,initial potential $V_1 = 10 \,V$,and final potential $V_2 = 20 \,V$.The increase in energy $\Delta U$ is given by $\Delta U = U_2 - U_1 = \frac{1}{2}C(V_2^2 - V_1^2)$.Substituting the values: $\Delta U = \frac{1}{2} 	imes (6 	imes 10^{-6}) 	imes (20^2 - 10^2)$.$\Delta U = 3 	imes 10^{-6} 	imes (400 - 100)$.$\Delta U = 3 	imes 10^{-6} 	imes 300 = 900 	imes 10^{-6} \,J$.$\Delta U = 9 	imes 10^{-4} \,J$.

If the potential of a capacitor having capacity of $6\,\mu F$ is increased from $10\,V$ to $20\,V$,then the increase in its energy will be:

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