Condenser A has a capacity of 15,mu F when it | vedclass

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(B) Charge on capacitor $A$ is given by $Q_1 = 15 	imes 10^{-6} 	imes 100 = 15 	imes 10^{-4}\,C$. Charge on capacitor $B$ is given by $Q_2 = 1 	im

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$800$

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(B) Charge on capacitor $A$ is given by $Q_1 = 15 	imes 10^{-6} 	imes 100 = 15 	imes 10^{-4}\,C$. Charge on capacitor $B$ is given by $Q_2 = 1 	imes 10^{-6} 	imes 100 = 10^{-4}\,C$. Capacity of capacitor $A$ after removing the dielectric is $C_A = \frac{15\,\mu F}{15} = 1\,\mu F$. When both capacitors are connected in parallel,their equivalent capacitance is $C_{eq} = C_A + C_B = 1\,\mu F + 1\,\mu F = 2\,\mu F$. The common potential $V$ is given by $V = \frac{Q_1 + Q_2}{C_{eq}} = \frac{15 	imes 10^{-4} + 1 	imes 10^{-4}}{2 	imes 10^{-6}} = \frac{16 	imes 10^{-4}}{2 	imes 10^{-6}} = 800\,V$.

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