If a slab of insulating material 4 times 10^{ | vedclass

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(B) Let the original separation between the plates be $d$ and the area of the plates be $A$. The original capacitance is $C = \frac{\epsilon_0 A}{d}$.

Vedclass · Accepted Answer

$8$

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(B) Let the original separation between the plates be $d$ and the area of the plates be $A$. The original capacitance is $C = \frac{\epsilon_0 A}{d}$.When a dielectric slab of thickness $t = 4 	imes 10^{-3} \ m$ is introduced,the new capacitance $C'$ becomes $C' = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$.To restore the capacity to its original value,the separation $d$ is increased by $\Delta d = 3.5 	imes 10^{-3} \ m$. The new separation is $d' = d + \Delta d$.The new capacitance is $C'' = \frac{\epsilon_0 A}{d' - t + \frac{t}{K}} = \frac{\epsilon_0 A}{d + \Delta d - t + \frac{t}{K}}$.Since $C'' = C$,we have $d = d + \Delta d - t + \frac{t}{K}$.This simplifies to $\Delta d = t(1 - \frac{1}{K})$.Substituting the values: $3.5 	imes 10^{-3} = 4 	imes 10^{-3} (1 - \frac{1}{K})$.$0.875 = 1 - \frac{1}{K} \implies \frac{1}{K} = 1 - 0.875 = 0.125$.$K = \frac{1}{0.125} = 8$.

If a slab of insulating material $4 \times 10^{-3} \ m$ thick is introduced between the plates of a parallel plate capacitor,the separation between plates has to be increased by $3.5 \times 10^{-3} \ m$ to restore the capacity to its original value. The dielectric constant of the material will be

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