Explain what would happen if in the capacitor given in Question 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,
(a) while the voltage supply remained connected.
(b) after the supply was disconnected.

(a)    Given,
Dielectric constant of mica sheet, k  = 6
Thickness ofmica sheet, t =3 mm = $3\times {10}^{-3}$ m
                                 $$\begin{gathered} \therefore \mathrm{Capacitance} \mathrm{of} \mathrm{the} \mathrm{plate}, C = {\mathrm{kC}}_0 \\ = 6 \times 18\times {10}^{-12}\mathrm{F} \\ = 108 \times {10}^{-12}\mathrm{F} \\ = 108\mathrm{p} \mathrm{F} \end{gathered}$$
Using the formula Q=CV
                           $$\begin{gathered} = 108 \times {10}^{-12} \times 100 \\ = 108 \times {10}^{-10} \end{gathered}$$
                        $\mathrm{Q} = 1.08 \times {10}^{-8}\mathrm{C}$

So, while voltage supply remained connected we have
$\mathrm{C} = 108 \mathrm{pF} \mathrm{and} \mathrm{Q} = 1.08 \times {10}^{-8}\mathrm{C}$

(b) After the supply was disconnected, the charge remains same
i.e.,                $\mathrm{q} = 1.8 \times {10}^{-9}\mathrm{C}$
and as                $\mathrm{Q} = \mathrm{CV}$
                       
                         $$\begin{gathered} \mathrm{V} = \frac{\mathrm{Q}}{\mathrm{C}} \\ = \frac{1.8 \times {10}^{-9}}{108 \times {10}^{-12}} \\ = 16.66 \mathrm{V} \end{gathered}$$

The charge remains constant i.e., Q = 1.08 x 10^–8 C after the supply was disconnected and the voltage will come down to 16.6 V.