A slab of material of dielectric constant k has the same area as the plates of the parallel plate capacitor but has thickness 4 3 d , where d is the separation of the plates. How is the capacitance

Question

A slab of material of dielectric constant k has the same area as the plates of the parallel plate capacitor but has thickness 4 3 d ​ , where d is the separation of the plates. How is the capacitance changed, when the slab is inserted between the plates? ( C = original capacitance)

TestHub · Accepted Answer

The original capacitance of the parallel plate capacitor is given by:

$C = \frac{ε _{0} A}{d}$

When a dielectric slab of thickness $t$ and dielectric constant $k$ is inserted between the plates, the new capacitance is:

$C^{'} = \frac{ε _{0} A}{d - t + \frac{t}{k}}$

Substituting $t = \frac{3 d}{4}$ into the expression:

$C^{'} = \frac{ε _{0} A}{d - \frac{3 d}{4} + \frac{3 d}{4 k}}$

$C^{'} = \frac{ε _{0} A}{\frac{d}{4} + \frac{3 d}{4 k}}$

$C^{'} = \frac{ε _{0} A}{\frac{d ( k + 3 )}{4 k}}$

$C^{'} = \frac{4 k}{k + 3} (\frac{ε _{0} A}{d})$

$C^{'} = \frac{4 k C}{k + 3}$

Answer: $\frac{4 k C}{( k + 3 )}$

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