(A) The initial capacitance of the air-filled capacitor is given by $C = \frac{\epsilon_0 A}{d}$, where $A = 12 \,cm^2$ and $d = 0.6 \,cm$.When a diel

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(A) The initial capacitance of the air-filled capacitor is given by $C = \frac{\epsilon_0 A}{d}$, where $A = 12 \,cm^2$ and $d = 0.6 \,cm$.When a dielectric slab of thickness $t = 0.6 \,cm$ and dielectric constant $K$ is inserted, and the plate separation is increased by $0.2 \,cm$ (new separation $d' = 0.6 + 0.2 = 0.8 \,cm$), the new capacitance is $C' = \frac{\epsilon_0 A}{d' - t + t/K}$.Given that $C = C'$, we have $\frac{\epsilon_0 A}{d} = \frac{\epsilon_0 A}{d' - t + t/K}$.Substituting the values: $\frac{1}{0.6} = \frac{1}{0.8 - 0.6 + 0.6/K}$.$0.6 = 0.2 + \frac{0.6}{K}$.$0.4 = \frac{0.6}{K}$.$K = \frac{0.6}{0.4} = 1.5$.

Class 12 Physics Electric Potential and Capacitance Effect of Dielectric Inside Capacitor

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$A$ capacitor has air as dielectric medium and two conducting plates of area $12 \,cm^2$ and they are $0.6 \,cm$ apart. When a slab of dielectric having area $12 \,cm^2$ and $0.6 \,cm$ thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2 \,cm$ to keep the capacitance same as in previous case. The dielectric constant of the slab is : (Given $\epsilon_0 = 8.834 \times 10^{-12} \,F/m$)

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Class 12

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

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$A$ capacitor of $10 \mu F$ capacitance,whose plates are separated by $10 \text{ mm}$ through air and each plate has an area of $4 \text{ cm}^2$,is now filled equally with two dielectric media of $K_1=2$ and $K_2=3$ respectively,as shown in the figure. If the new force between the plates is $8 \text{ N}$,the supply voltage is . . . . . . $V$.

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While a capacitor remains connected to a battery and a dielectric slab is inserted between the plates,then:

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