Question

A 2400 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?

Answer 1

Step-by-step explanation:

It is given that,

Capacitance,
`C=2400\ pF=2.4* 10^(-9)\ F`

Potential difference, V = 6.4 volts

Initial charge,
`Q_1=CV`

`Q_1=2.4* 10^(-9)\ F* 6.4\ V`

`Q_1=1.53* 10^(-8)\ C`

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,
`C'=2.4* 10^(-9)\ F* 7=1.68* 10^(-8)\ F`

Final charge,
`Q_2=1.68* 10^(-8)\ F* 6.4\ V`

`Q_2=1.07* 10^(-7)\ C`

`Q_2-Q_1=1.07* 10^(-7)\ C-1.53* 10^(-8)\ C=9.17* 10^(-8)\ C`

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