Question: (II) A 3500-pF air-gap capacitor is connected to a 32-V battery. If a piece of mica is placed between the plates, how much charge will flow from the battery?

The capacitance of a capacitor increases if the dielectric constant of the medium between the plates increases. When the capacitance increases, more charge is needed to maintain the same potential difference between the plates.

The voltage of the battery is,\(V = 32\;{\rm{V}}\)

The capacitance of the capacitor is, \(C = 3500\;{\rm{pF}}\).

The dielectric constant of mica is \(K = 7\).

The initial charge on the capacitor is,

\(Q = CV\)

Initially, there is air between the plates. The dielectric constant increases when the mica is inserted between the plates.

The new capacitance of the capacitor is,

\(C' = KC\)

Now, the charge on the capacitor is,

\(\begin{aligned}{c}Q' &= C'V\\ &= KCV\end{aligned}\).

The extra charge flows from the battery is,

\(\begin{aligned}{c}\Delta Q &= Q' - Q\\ &= KCV - CV\\ &= \left( {K - 1} \right)CV\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{c}\Delta Q &= \left( {7 - 1} \right)\left( {3500\;{\rm{pF}}} \right)\left( {32\;{\rm{V}}} \right)\\ &= 672000\;{\rm{pC}}\\ &= 6.72 \times {10^{ - 7}}\;{\rm{C}}\end{aligned}\)

Thus, the charge flows from the battery is \(6.72 \times {10^{ - 7}}\;{\rm{C}}\).