91Ó°ÊÓ

The capacitance of a capacitor relies on the area of capacitor plates and the separation between the plates. The value of capacitance will increase 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 the plate and d is the separation between plates.

The diameter of the pie pans is,\(D = 9\;{\rm{in}}\)

The separation of the pans is,\(d = 4\;{\rm{cm}} = 0.04\;{\rm{m}}\).

The potential of the battery is, \(V = 9\;{\rm{V}}\).

The area of the pans is,

\(A = \pi \frac{{{D^2}}}{4}\)

The capacitance of the capacitor is given as:

\(\begin{aligned}{c}C &= \frac{{{\varepsilon _0}A}}{d}\\ &= \frac{{{\varepsilon _0}\left( {\pi \frac{{{D^2}}}{4}} \right)}}{d}\\ &= \frac{{{\varepsilon _0}\pi {D^2}}}{{4d}}\end{aligned}\).

Substitute the values in the above expression.

\(\begin{aligned}{c}C &= \frac{{\left( {8.854 \times {{10}^{ - 12}}\;{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {{\rm{m}}^{\rm{2}}}} \right) \times 3.14 \times {{\left( {9\;{\rm{in}} \times \frac{{0.254\;{\rm{m}}}}{{1\;{\rm{in}}}}} \right)}^2}}}{{4 \times \left( {0.04\;{\rm{m}}} \right)}}\\ &= 9.08 \times {10^{ - 12}}\;{\rm{F}}\\ \approx 9\;{\rm{pF}}\end{aligned}\)

Hence, the capacitance is \(9\;{\rm{pF}}\).

The charge on each plate is given as:

\(Q = CV\)

Substitute the values in the above expression.

\(\begin{aligned}{c}Q &= \left( {9.0 \times {{10}^{ - 12}}\;{\rm{F}}} \right) \times \left( {9\;{\rm{V}}} \right)\\ &= 81 \times {10^{ - 12}}\;{\rm{C}}\\ \approx 81\;{\rm{pC}}\end{aligned}\)

Thus, the charge on each plate is \(81\;{\rm{pC}}\).

The electric field is the same everywhere inside the plates.

The electric field at any point between the plates is given as:

\(E = \frac{V}{d}\)

Substitute the values in the above expression.

\(\begin{aligned}{c}E &= \frac{{9\;{\rm{V}}}}{{0.04\;{\rm{m}}}}\\ &= 225\;{\rm{V/m}}\end{aligned}\)

Thus, the electric field halfway between the plates is 225 V/m.

The work done by the battery to charge the capacitor is equal to the potential energy stored in the capacitor.

Then the work done by the battery is,

\(W = \frac{1}{2}QV\)

Substitute the values in the above equation.

\(\begin{aligned}{c}W &= \frac{1}{2} \times \left( {81 \times {{10}^{ - 12}}\;{\rm{C}}} \right) \times \left( {9\;{\rm{V}}} \right)\\ &= 3.65 \times {10^{ - 10}}\;{\rm{J}}\\ \approx 4 \times {10^{ - 10}}\;{\rm{J}}\end{aligned}\)

Thus, the work done by the battery to charge the capacitor is \(4 \times {10^{ - 10}}\;{\rm{J}}\).

In the presence of dielectric material, the capacitance is given by,

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

Here, K is the dielectric constant of the material.

Since the value of\(K > 1\), then the value of the capacitor will increase.

Since the charge stored on the plates depends on the value of capacitance, then with the increase in capacitance, the charge will also increase.

Since the work done by the battery depends on the charge stored on the plates, the work done will also increase with the increase in charge.

Since the electric field between the plates is independent of the capacitance and depends only on the separation between the plates and voltage of battery, then the value of electric field will remain the same after inserting the dielectric material.