Explanation:
It is given that,
Capacitance, [tex]C=2400\ pF=2.4\times 10^{-9}\ F[/tex]
Potential difference, V = 6.4 volts
Initial charge, [tex]Q_1=CV[/tex]
[tex]Q_1=2.4\times 10^{-9}\ F\times 6.4\ V[/tex]
[tex]Q_1=1.53\times 10^{-8}\ C[/tex]
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, [tex]C'=2.4\times 10^{-9}\ F\times 7=1.68\times 10^{-8}\ F[/tex]
Final charge, [tex]Q_2=1.68\times 10^{-8}\ F\times 6.4\ V[/tex]
[tex]Q_2=1.07\times 10^{-7}\ C[/tex]
[tex]Q_2-Q_1=1.07\times 10^{-7}\ C-1.53\times 10^{-8}\ C=9.17\times 10^{-8}\ C[/tex]
So, when mica sheet is introduced, more charge will flow out of the battery. Hence, this is the required solution.
