When the capacitor is connected to the voltage, a charge Q is stored on its plates. Calling [tex] C_0 [/tex] the capacitance of the capacitor in air, the charge Q, the capacitance [tex] C_0 [/tex] and the voltage ([tex] V_0=391 V [/tex]) are related by
[tex] C_0 =\frac{Q}{V_0} [/tex] (1)
when the source is disconnected the charge Q remains on the capacitor.
When the space between the plates is filled with mica, the capacitance of the capacitor increases by a factor 5.4 (the permittivity of the mica compared to that of the air):
[tex] C=k C_0 = 5.4 C_0 [/tex]
this is the new capacitance. Since the charge Q on the plates remains the same, by using eq. (1) we can find the new voltage across the capacitor:
[tex] V=\frac{Q}{C}=\frac{Q}{5.4 C_0} [/tex]
And since [tex] Q=C_0 V_0 [/tex], substituting into the previous equation, we find:
[tex] V=\frac{C_0 V_0}{5.4 C_0}=\frac{V_0}{5.4}=\frac{391 V}{5.4}=72.4 V [/tex]
