The capacity of a parallel plate condenser is C 0 . If a dielectric of relative permittivity ε r and thickness equal to one fourth the plate separation is placed between the plates, then its capac

Question

The capacity of a parallel plate condenser is C 0 ​ . If a dielectric of relative permittivity ε r ​ and thickness equal to one fourth the plate separation is placed between the plates, then its capacity becomes C . Then value of C 0 ​ C ​ will be-

TestHub · Accepted Answer

To refine the solution, we first define the initial capacitance $C_{0} = \frac{ε _{0} A}{d}$ . When a dielectric of thickness $t = \frac{d}{4}$ and relative permittivity $ε_{r}$ is inserted, the capacitor can be treated as two capacitors in series: one with dielectric and thickness $t$ , and one with air and thickness $d - t$ .

The capacitance of the part with dielectric is $C_{1} = \frac{ε _{r} ε _{0} A}{t} = \frac{ε _{r} ε _{0} A}{d /4} = \frac{4 ε _{r} ε _{0} A}{d}$ .

The capacitance of the part with air is $C_{2} = \frac{ε _{0} A}{d - t} = \frac{ε _{0} A}{d - d /4} = \frac{ε _{0} A}{3 d /4} = \frac{4 ε _{0} A}{3 d}$ .

For capacitors in series, the equivalent capacitance $C$ is given by:

$\frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}}$

$\frac{1}{C} = \frac{d}{4 ε _{r} ε _{0} A} + \frac{3 d}{4 ε _{0} A} = \frac{d}{4 ε _{0} A} (\frac{1}{ε _{r}} + 3)$

$\frac{1}{C} = \frac{d}{4 ε _{0} A} (\frac{1 + 3 ε _{r}}{ε _{r}})$

$C = \frac{4 ε _{0} A}{d} (\frac{ε _{r}}{1 + 3 ε _{r}})$

Now, we need to find the ratio $\frac{C}{C _{0}}$ :

$\frac{C}{C _{0}} = \frac{\frac{4 ε _{0} A}{d} ( \frac{ε _{r}}{1 + 3 ε _{r}} )}{\frac{ε _{0} A}{d}}$

$\frac{C}{C _{0}} = \frac{4 ε _{r}}{1 + 3 ε _{r}}$

The final answer is $\frac{4 ε _{r}}{3 ε _{r} + 1}$ .

Physics - Capacitor Question with Solution | TestHub

Options:

Doubts & Discussion

JEE Main (2026)