Question

Question: (I) What is the capacitance of a pair of circular plates with a radius of 5.0 cm separated by 2.8 mm of mica?

Short answer

The capacitance of the pair of circular plates is \(1.7 \times {10^{ - 10}}\;{\rm{F}}\).

Step-by-step solution

Understanding the effect of dielectric on capacitance

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.

Given data

The radius of the circular plates is,\(r = 5.0\;{\rm{cm}} = 0.05\;{\rm{m}}\).

The separation between the plates is,\(d = 2.8\;{\rm{mm}} = 2.8 \times {10^{ - 3}}\;{\rm{m}}\).

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

Determination of the capacitance

The area of the plates is,

\(A = \pi {r^2}\)

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

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

Substitute the values in the above expression.

\(\begin{aligned}{c}C &= 7 \times \left( {8.854 \times {{10}^{ - 12}}\;{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {{\rm{m}}^{\rm{2}}}} \right) \times \frac{{3.14 \times {{\left( {0.05\;{\rm{m}}} \right)}^2}}}{{2.8 \times {{10}^{ - 3}}\;{\rm{m}}}}\\ &= 1.7 \times {10^{ - 10}}\;{\rm{F}}\end{aligned}\)

Thus, the capacitance of the pair of circular plates is \(1.7 \times {10^{ - 10}}\;{\rm{F}}\).