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Explain what would happen if in the capacitor given in 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, While the voltage supply remained connected.

$\begin{array}{1 1} 10 V \\ 100\;V \\ 1000\;V \\ 50\;V\end{array} $

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A)
Solution :
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} $
New capacitance, $C1 = kC = 6 × 1.771 × 10^{−11} F = 106 pF$
New charge, $q_1 = C_1V = 106 × 100 pC = 1.06 × 10^{–8 }\;C$
Potential across the plates remains 100 V.
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