Question

The figure below shows a parallel-plate capacitor with a plate area A = 6.67 cm2 and plate separation d = 4.62 mm. The top half of the gap is filled with material of dielectric constant κ1 = 7.50; the bottom half is filled with material of dielectric constant κ2 = 14.5. What is the capacitance

F

Answer 1

Concepts and reason

The concepts used to solve this problem are capacitance of the capacitor when dielectric is placed in between the plates, series capacitance.

Initially, obtain the expression for the capacitance of the parallel plate capacitor by using the expression for the equivalent capacitance of the capacitor plates connected in series and expression for the capacitance of a capacitor when dielectric is placed in between the plates.

Finally, substitute the values in the expression obtained in step 1 to get the value of the capacitance of the parallel plate capacitor.

Fundamentals

The capacitance of the capacitor when the dielectric is placed in between the plates is as follows:

$C = \frac{{k{\varepsilon _0}A}}{d}$

Here, k is the dielectric constant, ${\varepsilon _0}$ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.

The equivalent capacitance of n capacitors when they are connected in series is as follows:

$\frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + .... + \frac{1}{{{C_n}}}$

Here, ${C_{eq}}$ is the equivalent capacitance of the series capacitors, ${C_1}$ is the capacitance of the first capacitor, ${C_2}$ is the capacitance of the second capacitor, and ${C_n}$ is the capacitance of the ${n^{th}}$ capacitor.

The capacitance of the first capacitor is,

${C_1} = \frac{{{k_1}{\varepsilon _0}A}}{{{d_1}}}$

Here, ${k_1}$ is the dielectric constant of the material that is filled in the top half of the parallel plate capacitor and ${d_1}$ is the distance up to which the top half of the dielectric material is filled between the plates.

The capacitance of the second capacitor is,

${C_2} = \frac{{{k_2}{\varepsilon _0}A}}{{{d_2}}}$

Here, ${k_2}$ is the dielectric constant of the material that is filled in the bottom half of the parallel plate capacitor and ${d_2}$ is the distance up to which the bottom half of the dielectric material is filled between the plates.

The equivalent capacitance of two capacitors when they are connected in series is,

$\frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}$

Substitute $\frac{{{k_1}{\varepsilon _0}A}}{{{d_1}}}$ for ${C_1}$ and $\frac{{{k_2}{\varepsilon _0}A}}{{{d_2}}}$ for ${C_2}$ .

$\begin{array}{c}\\\frac{1}{{{C_{eq}}}} = \frac{1}{{\left( {\frac{{{k_1}{\varepsilon _0}A}}{{{d_1}}}} \right)}} + \frac{1}{{\left( {\frac{{{k_2}{\varepsilon _0}A}}{{{d_2}}}} \right)}}\\\\ = \frac{1}{{{\varepsilon _0}A}}\left( {\frac{{{d_1}}}{{{k_1}}} + \frac{{{d_2}}}{{{k_2}}}} \right)\\\end{array}$

The capacitance of the capacitor is,

$\frac{1}{{{C_{eq}}}} = \frac{1}{{{\varepsilon _0}A}}\left( {\frac{{{d_1}}}{{{k_1}}} + \frac{{{d_2}}}{{{k_2}}}} \right)$

Substitute $\left( {\frac{d}{2}} \right)$ for ${d_1}$ and ${d_2}$ .

$\begin{array}{c}\\\frac{1}{{{C_{eq}}}} = \frac{1}{{{\varepsilon _0}A}}\left( {\frac{{\left( {\frac{d}{2}} \right)}}{{{k_1}}} + \frac{{\left( {\frac{d}{2}} \right)}}{{{k_2}}}} \right)\\\\ = \frac{d}{{2{\varepsilon _0}A}}\left( {\frac{1}{{{k_1}}} + \frac{1}{{{k_2}}}} \right)\\\\{C_{eq}} = \frac{{2{\varepsilon _0}A}}{d}\left( {\frac{{{k_1}{k_2}}}{{{k_1} + {k_2}}}} \right)\\\end{array}$

Substitute $8.85 \times {10^{ - 12}}\;{{\rm{C}}^2}/{\rm{N}} \cdot {{\rm{m}}^2}$ for ${\varepsilon _0}$ , $6.67\;{\rm{c}}{{\rm{m}}^2}$ for A, 4.62 mm for d, 7.50 for ${k_1}$ , and 14.5 for ${k_2}$ .

$\begin{array}{c}\\{C_{eq}} = \frac{{2\left( {8.85 \times {{10}^{ - 12}}\;{{\rm{C}}^2}/{\rm{N}} \cdot {{\rm{m}}^2}} \right)\left( {6.67\;{\rm{c}}{{\rm{m}}^2}\left( {\frac{{1\;{{\rm{m}}^2}}}{{{{10}^4}\;{\rm{c}}{{\rm{m}}^2}}}} \right)} \right)}}{{4.62\;{\rm{mm}}\left( {\frac{{1\;{\rm{m}}}}{{{{10}^3}\;{\rm{m}}}}} \right)}}\left( {\frac{{\left( {7.50} \right)\left( {14.5} \right)}}{{7.5 + 14.5}}} \right)\\\\ = 126.32 \times {10^{ - 11}}\;{\rm{F}}\\\\ = 1263.2\;{\rm{pF}}\\\end{array}$

Ans:

The capacitance of the capacitor is 1263.2 pF.