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(A) The energy stored in a capacitor connected to a battery is given by $U = \frac{1}{2} C V^2$,where $C = \frac{K \epsilon_0 A}{d}$.Since the battery

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increase by $50 \%$

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(A) The energy stored in a capacitor connected to a battery is given by $U = \frac{1}{2} C V^2$,where $C = \frac{K \epsilon_0 A}{d}$.Since the battery remains connected,the potential difference $V$ remains constant.Initial energy $U_1 = \frac{1}{2} (K_1 C_0) V^2$,where $C_0 = \frac{\epsilon_0 A}{d}$.Final energy $U_2 = \frac{1}{2} (K_2 C_0) V^2$.The percentage change in energy is given by $\frac{U_2 - U_1}{U_1} 	imes 100 \%$.Substituting the values: $\frac{\frac{1}{2} K_2 C_0 V^2 - \frac{1}{2} K_1 C_0 V^2}{\frac{1}{2} K_1 C_0 V^2} 	imes 100 \% = \frac{K_2 - K_1}{K_1} 	imes 100 \%$.Given $K_1 = 10$ and $K_2 = 15$,the change is $\frac{15 - 10}{10} 	imes 100 \% = \frac{5}{10} 	imes 100 \% = 50 \%$.Thus,the energy increases by $50 \%$.

$A$ parallel plate capacitor filled with a medium of dielectric constant $10$ is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant $15$. Then the energy of the capacitor will ......................

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