# Explain what would happen if in the capacitor given in Exercise 2.8,

Question:

Explain what would happen if in the capacitor given in Exercise 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

Solution:

(a) Dielectric constant of the mica sheet, k = 6

Initial capacitance, $C=1.771 \times 10^{-11} \mathrm{~F}$

New capacitance, $C^{\prime}=k C=6 \times 1.771 \times 10^{-11}=106 \mathrm{pF}$

Supply voltage, $V=100 \mathrm{~V}$

New charge, $q^{\prime}=C^{\prime} V=6 \times 1.771 \times 10^{-9}=1.06 \times 10^{-8} \mathrm{C}$

Potential across the plates remains $100 \mathrm{~V}$.

(b) Dielectric constant, $k=6$

Initial capacitance, $C=1.771 \times 10^{-11} \mathrm{~F}$

New capacitance, $C^{\prime}=k C=6 \times 1.771 \times 10^{-11}=106 \mathrm{pF}$

If supply voltage is removed, then there will be no effect on the amount of charge in the plates.

Charge $=1.771 \times 10^{-9} \mathrm{C}$

Potential across the plates is given by,

$\therefore V^{\prime}=\frac{q}{C^{\prime}}$

$=\frac{1.771 \times 10^{-9}}{106 \times 10^{-12}}$

$=16.7 \mathrm{~V}$