The capacitance of a capacitor becomes 6 7 times the original value if dielectric slab of thickness t = 3 2 d is introduced between the plates, where 'd' is the distance of separation be

Question

The capacitance of a capacitor becomes 6 7 ​ times the original value if dielectric slab of thickness t = 3 2 d ​ is introduced between the plates, where 'd' is the distance of separation between the plates. The dielectric constant of the slab is

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The initial capacitance of the capacitor is given by:

$C_{0} = \frac{ϵ _{0} A}{d}$

When a dielectric slab of thickness $t$ and dielectric constant $K$ is introduced, the new capacitance becomes:

$C = \frac{ϵ _{0} A}{d - t + \frac{t}{K}}$

Substituting $t = \frac{2 d}{3}$ into the expression:

$C = \frac{ϵ _{0} A}{d - \frac{2 d}{3} + \frac{2 d}{3 K}}$

$C = \frac{ϵ _{0} A}{\frac{d}{3} + \frac{2 d}{3 K}} = \frac{3 ϵ _{0} A}{d ( 1 + \frac{2}{K} )}$

$C = \frac{3 C _{0}}{1 + \frac{2}{K}}$

Given that $C = \frac{7}{6} C_{0}$ , we can equate the two expressions:

$\frac{7}{6} = \frac{3}{1 + \frac{2}{K}}$

$7 (1 + \frac{2}{K}) = 18$

$7 + \frac{14}{K} = 18$

$\frac{14}{K} = 11$

$K = \frac{14}{11}$

Answer: $\frac{14}{11}$

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