A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness ((3/4)) d, where d is the separation of the plates. The ratio of the capacitance C (in the presence of the dielectric) to the capacitance C0 (in the absence of the dielectric) is

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 ((3/4)) d, where d is the separation of the plates. The ratio of the capacitance C (in the presence of the dielectric) to the capacitance C0 (in the absence of the dielectric) is (A)  (3K/K+4)  (B)  (3/4)K  (C)  (4K/K+3)K  (D)  (4/3)K

Tardigrade Admin · Accepted Answer

The capacitance of a parallel plate capacitor in the absence of the dielectric is

C_{0} = \frac{ε _{0} A}{d}

If a dielectric slab is partially filled between the plates of capacitor

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

C = \frac{ε _{0} A}{( d - \frac{3}{4} d ) + ( \frac{3 d}{4 K} )}

C = \frac{ε _{0} A}{\frac{d}{4} + \frac{3 d}{4 K}}

= \frac{4 K ε _{0} A}{d ( K + 3 )}

∴ \frac{C}{C _{0}} = \frac{4 K ε _{0} A}{d ( K + 3 )} \times \frac{d}{ε _{0} A}

= \frac{4 K}{( K + 3 )}

$\frac{3 K}{K + 4}$

$\frac{3}{4} K$

$\frac{4 K}{K + 3} K$

$\frac{4}{3} K$

K+43K​

43​K

K+34K​K

34​K

$\frac{3 K}{K + 4}$

$\frac{3}{4} K$

$\frac{4 K}{K + 3} K$

$\frac{4}{3} K$