# 2.9) 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.

P Pankaj Sanodiya

Given,

The dielectric constant of the inserted mica sheet = 6

The thickness of the sheet = 3mm

Supply Voltage $V = 100V$

Initial Capacitance = $C_{initial}=1.771 *10^{-11}F$

Final Capacitance = $KC_{initial}=6*1.771 *10^{-11}F=106*10^{-12}F$

Final Charge on the capacitor = $Q_{final}=C_{final}V=106*10^{-12}*100=106*10^{-10}C$

Hence on inserting the sheet charge on each plate changes to $106*10^{-10}C$.

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