A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness ( 4/5 ) d , where d is the separation of the plates. The capacitance in th

Question

A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness ( 4/5 ) d , where d is the separation of the plates. The capacitance in the presence and absence of dielectric are C and C 0 ​ respectively. The ratio ( C / C 0 ​ ) is

TestHub · Accepted Answer

The capacitance of a parallel plate capacitor without a dielectric is given by:

$C_{0} = \frac{ε _{0} A}{d}$

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

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

Given $t = \frac{4}{5} d$ , substituting this value yields:

$C = \frac{ε _{0} A}{d - \frac{4}{5} d + \frac{4 d}{5 K}}$

$C = \frac{ε _{0} A}{\frac{d}{5} + \frac{4 d}{5 K}}$

$C = \frac{ε _{0} A}{\frac{d ( K + 4 )}{5 K}}$

$C = \frac{5 K ε _{0} A}{d ( K + 4 )}$

Taking the ratio of $C$ to $C_{0}$ gives:

$\frac{C}{C _{0}} = \frac{5 K}{K + 4}$

Answer: $\frac{5 K}{K + 4}$

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