In a parallel plate capacitor, each plate having an area $6\times {10}^{-3}{m}^2$ and distance between the plates $3\text{mm}$ is connected to a 100 V supply. If a $3\text{mm}$ thick mica sheet (of dielectric constant = 6) is inserted between the plates, calculate the capacitance and charge on each plate
A) While the voltage supply remained connected

In a parallel plate capacitor, each plate having an area $6\times {10}^{-3}{m}^2$ and distance between the plates $3\text{mm}$ is connected to a 100 V supply. If a $3\text{mm}$ thick mica sheet (of dielectric constant = 6) is inserted between the plates, calculate the capacitance and charge on each plate
B) after the supply was disconnected

A) Step 1: Find the capacitance of the capacitor.
Given,
Dielectric constant of mica sheet, k = 6
Thickness of mica sheet, $t=3\text{mm}=3\times {10}^{-3}m$
$\therefore$ Capacitance of the plate, $C=k{C}_0$
$=6\times 17.7\times {10}^{-12}F$
$\approx 106\times {10}^{-12}F$
$\approx 106pF$
Step 2: Find the charge of the capacitor.
using the formula Q = CV
$=106\times {10}^{-12}\times 100$
$=106\times {10}^{-10}C$
So, while voltage supply remained connected, we have
$C=106pFQ=1.06\times {10}^{-8}C$.

B) Given,
Dielectric constant of mica sheet, k = 6
Thickness of mica sheet, $t=3\text{mm}=3\times {10}^{-3}m$
$\therefore$ Capacitance of the plate,
$C=k{C}_0$
$=6\times 17.7\times {10}^{-12}F$
$\approx 106\times {10}^{-12}F$
$\approx 106pF$
After the supply was disconnected, the charge remains same and as
$Q=CV$
$V=\frac{Q}{C}=\frac{1.77\times {10}^{-9}C}{106\times {10}^{-12}}$
$=16.7V$
The charge remains constant i.e., $Q=1.77\times {10}^{-9}C$ after the supply was disconnected and the voltage will come down to 16.7 V.