# 2.9) Explain what would happen if in the capacitor given, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,(b) after the supply was disconnected.

If a 3mm mica sheet is inserted between plates of the capacitor after disconnecting it from the power supply, the voltage across the capacitor be changed. Since the charge on the capacitor can not go anywhere, if we change the capacitance (which we are doing by inserting mica sheet here), the voltage across the capacitor has to be adjusted accordingly.

As obtained from question number 8 charge on each plate of the capacitor is $1.771*10^{-9}C$

$V_{final}=\frac{Q}{C_{final}}=\frac{1.771\times10^{-9}}{106\times 10^{-12}}=16.7V$

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