In a parallel plate capacitor with air between the plates,each plate has an area of $6 \times 10^{-3} \, m^{2}$ and the distance between the plates is $3 \, mm$. The capacitance of the capacitor is $17.71 \, pF$. If this capacitor is connected to a $100 \, V$ supply,and a $3 \, mm$ thick mica sheet (of dielectric constant $k = 6$) is inserted between the plates,calculate the new capacitance,charge,and potential difference in the following cases:
$(a)$ While the voltage supply remains connected.
$(b)$ After the supply is disconnected.

(A) Dielectric constant of the mica sheet,$k = 6$.
If the voltage supply remains connected,the potential difference between the plates remains constant. Supply voltage,$V = 100 \, V$.
Initial capacitance,$C = 17.71 \, pF = 1.771 \times 10^{-11} \, F$.
New capacitance,$C_{1} = k \cdot C = 6 \times 17.71 \, pF = 106.26 \, pF$.
New charge,$q_{1} = C_{1} \cdot V = 106.26 \times 10^{-12} \, F \times 100 \, V = 1.0626 \times 10^{-8} \, C$.
The potential across the plates remains $100 \, V$.
$(b)$ Dielectric constant,$k = 6$.
Initial charge,$q = C \cdot V = 17.71 \times 10^{-12} \, F \times 100 \, V = 1.771 \times 10^{-9} \, C$.
New capacitance,$C_{1} = k \cdot C = 6 \times 17.71 \, pF = 106.26 \, pF$.
If the supply is disconnected,the charge remains constant,$q = 1.771 \times 10^{-9} \, C$.
The new potential across the plates is given by $V_{1} = \frac{q}{C_{1}} = \frac{1.771 \times 10^{-9} \, C}{106.26 \times 10^{-12} \, F} \approx 16.67 \, V$.