Explanation

Initially, the capacitor is connected to a battery, so the potential difference across the plates remains constant.

Let the original plate separation be

$$ d = 5\,mm $$

Initial capacitance is

$$ C_0 = \frac{\varepsilon_0 A}{d} $$

After inserting a mica sheet of thickness $t = 2\,mm$ and dielectric constant $k$, the capacitor becomes a combination of two layers:

• Mica layer of thickness $2\,mm$ • Air gap of thickness $5 - 2 = 3\,mm$

The effective separation is given by

$$ d_{\text{eff}} = \frac{t}{k} + (d - t) $$

So,

$$ d_{\text{eff}} = \frac{2}{k} + 3 $$

New capacitance becomes

$$ C = \frac{\varepsilon_0 A}{\dfrac{2}{k} + 3} $$

Since the battery is connected, charge is proportional to capacitance. Given that the capacitor draws $25\%$ more charge,

$$ C = 1.25\,C_0 $$

Substituting,

$$ \frac{\varepsilon_0 A}{\dfrac{2}{k} + 3} = 1.25 \times \frac{\varepsilon_0 A}{5} $$

Cancelling $\varepsilon_0 A$,

$$ \frac{1}{\dfrac{2}{k} + 3} = \frac{1.25}{5} $$

$$ \frac{1}{\dfrac{2}{k} + 3} = \frac{1}{4} $$

Taking reciprocal,

$$ \frac{2}{k} + 3 = 4 $$

$$ \frac{2}{k} = 1 $$

$$ k = 2 $$

Therefore, the dielectric constant of mica is

$$ k = 2.0 $$

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