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
(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,
$\therefore V^{\prime}=\frac{q}{C^{\prime}}$
$=\frac{1.771 \times 10^{-9}}{106 \times 10^{-12}}$
$=16.7 \mathrm{~V}$
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