For charging the capacitance of a given parallel plate capacitor, a dielectric material of constant K is used, which has same area as the plates of the capacitor. The thickness of dielectric slab is 3d4 , where d is the separation between the plates of parallel plate capacitor. The new capacitance C1 in terms of original capacitance C0 is given by the following relation.

C0=ε0AdC1=ε0Ad−3d41−1K=ε0A4d−3d1−1K=4ε0Ad−3dK4KK+3C0

For charging the capacitance of a given parallel plate capacitor, a dielectric material of constant K is used, which has same area as the plates of the capacitor. The thickness of dielectric slab is 3d4 , where d is the separation between the plates of parallel plate capacitor. The new capacitance C1 in terms of original capacitance C0 is given by the following relation.

C0=ε0AdC1=ε0Ad−3d41−1K=ε0A4d−3d1−1K=4ε0Ad−3dK4KK+3C0

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For charging the capacitance of a given parallel plate capacitor, a dielectric material of constant K is used, which has same area as the plates of the capacitor. The thickness of dielectric slab is $\frac{3 d}{4}$ , where d is the separation between the plates of parallel plate capacitor. The new capacitance C¹ in terms of original capacitance $C_{0}$ is given by the following relation.

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$C^{1} = (\frac{3 + K}{4 K}) C_{0}$

$C^{1} = \frac{(4 + K) C_{0}}{3}$

$C^{1} = \frac{4 K C_{0}}{K + 3}$

$C^{1} = \frac{4 C_{0}}{3 + K}$

answer is C.

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Detailed Solution

$C_{0} = \frac{ε_{0} A}{d}$

$C^{1} = \frac{ε_{0} A}{d - \frac{3 d}{4} (1 - \frac{1}{K})} = \frac{ε_{0} A}{4 d - 3 d (1 - \frac{1}{K})} = \frac{4 ε_{0} A}{d - \frac{3 d}{K}}$

$(\frac{4 K}{K + 3}) C_{0}$

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