91Ó°ÊÓ

The capacitance of a capacitor relies on the area of capacitor plates and separation between the plates. The value of capacitor increases when a dielectric material is inserted between the plates.

The expression for the capacitor is given as:

\(C = {\varepsilon _0}\frac{A}{d}\) … (i)

Here,\({\varepsilon _0}\)is the permittivity of free space, A is the area of plate and d is the separation between plates.

The charge on the capacitor remains the same after disconnecting the battery.

The voltage of the battery is,\(V = 21.0\;{\rm{V}}\)

The dielectric constant of paraffin is \(K = 2.2\).

The charge on the capacitor is,

\(Q = CV\)

After disconnecting from the battery, the charge on the plates remains same. Initially, there is air between the plates.

The capacitance of the capacitor when paraffin is inserted between the plates is,

\(\begin{aligned}{l}C' &= K{\varepsilon _0}\frac{A}{d}\\C' &= KC\end{aligned}\)

As the charge remains the same then,

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

Substitute the values in the above expression.

\(\begin{aligned}{c}V' &= \frac{{21.0\;{\rm{V}}}}{{2.2}}\\ &= 9.54\;{\rm{V}}\\ \approx 9.5\;{\rm{V}}\end{aligned}\)

Thus, the voltage between the plates is 9.5 V.