Question: (I) What is the capacitance of two square parallel plates 6.6 cm on a side that are separated by 1.8 mm of paraffin?

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:

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

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

The side of the square plate is,\(s = 6.6\;{\rm{cm}} = 0.066\;{\rm{m}}\).

The separation between the plates is, \(d = 1.8\;{\rm{mm}} = 1.8 \times {10^{ - 3}}\;{\rm{m}}\).

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

The area of the plates is,

\(A = {s^2}\)

From equation (i), the capacitance of the capacitor is,

\(C = K{\varepsilon _0}\frac{{{s^2}}}{d}\)

Substitute the values in the above expression.

\(\begin{aligned}{c}C &= 2.2 \times \left( {8.854 \times {{10}^{ - 12}}\;{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {{\rm{m}}^{\rm{2}}}} \right) \times \frac{{{{\left( {0.066\;{\rm{m}}} \right)}^2}}}{{1.8 \times {{10}^{ - 3}}\;{\rm{m}}}}\\ &= 4.7 \times {10^{ - 11}}\;{\rm{F}}\end{aligned}\)

Thus, the capacitance of the capacitor is \(4.7 \times {10^{ - 11}}\;{\rm{F}}\).