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, $3\; mm$ 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.
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, $3\; mm$ 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.
$(a)$ Dielectric constant of the mica sheet, $k=6$
If voltage supply remained connected, voltage between two plates will be constant. Supply voltage, $V =100 \,V$ Initial capacitance, $C =1.771 \times 10^{-11}\, F$
New capacitance, $C _{1}=k \,C =6 \times 1.771 \times 10^{-11} \,F =106\, pF$
New charge, $q_{1}=C_{1} \,V=106 \times 100 \,pC =1.06 \times 10^{-8}\, C$
Potential across the plates remains $100\, V$
$(b)$ Dielectric constant, $k=6$ Initial capacitance, $C =1.771 \times 10^{-11}\, F$
New capacitance, $C _{1}=k\, C =6 \times 1.771 \times 10^{-11} F =106 \,pF$
If supply voltage is removed, then there will be constant amount of charge in the plates. Charge $=1.771 \times 10^{-9}\,C$
Potential across the plates is given by,
$V_{1}=\frac{q}{C_{1}}=\frac{1.771 \times 10^{-9}}{106 \times 10^{-12}}=16.7 \,V$
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