The plates of a parallel plate capacitor are separated by a distance 'd' with air as the medium between them. A dielectric slab of dielectric constant 3 is introduced between the plates so a

Question

The plates of a parallel plate capacitor are separated by a distance 'd' with air as the medium between them. A dielectric slab of dielectric constant 3 is introduced between the plates so as to increase the capacity by 50% . The thickness of the dielectric slab is

TestHub · Accepted Answer

The initial capacitance of the parallel plate capacitor with air as the medium is given by:

$C_{0} = \frac{ε _{0} A}{d}$

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

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

Given that the capacity increases by $50%$ , the new capacitance is:

$C = C_{0} + 0.5 C_{0} = \frac{3}{2} C_{0}$

Substituting the expressions for $C$ and $C_{0}$ :

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

$\frac{1}{d - t + \frac{t}{3}} = \frac{3}{2 d}$

$2 d = 3 (d - t + \frac{t}{3})$

$2 d = 3 (d - \frac{2 t}{3})$

$2 d = 3 d - 2 t$

$2 t = d$

$t = \frac{d}{2}$

Answer: $\frac{d}{2}$

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