A2400 pf air-gap capacitor is connected to a 6.4 v battery. if a piece of mica is placed between the plates, how much charge will flow from the battery?

Explanation:

It is given that,

Capacitance,

Potential difference, V = 6.4 volts

Initial charge,

When the mica is introduced between the plates of capacitor, the capacitance increases by a factor of k. The dielectric constant of mica lies in between 6 - 8. Let k = 7 (say)

New capacitance,

Final charge,

So, when mica sheet is introduced, more charge will flow out of the battery. Hence, this is the required solution.