Question

Six parallel-plate capacitors of identical plate separation have different plate areas A, different capacitances C,and different dielectrics filling the space between theplates.

Part A. Rank the following capacitors on thebasis of the dielectric constant of the material between theplates.

Rank from largest tosmallest.

1. A = 4 cm^2    C = 2 nF

2. A = 1cm^2     C = 1 nF

3. A = 2cm^2     C = 8 nF

4. A = 8 cm^2    C = 2 nF

5. A = 4 cm^2    C = 1 nF

6. A = 2cm^2     C = 4 nF

All of the capacitors from Part A arenow attached to batteries with the same potential difference.

Rank the capacitors on the basis of thecharge stored on the positive plate.

Rank from largest to smallest.

Answer 1

Concepts and reason

The concepts required to solve the problem are the capacitance of parallel plate capacitors, the charge stored in the capacitor.

Derive an equation for the dielectric constant of the parallel plate capacitor from the capacitance of the capacitor. Take all the constant values as proportionality constant and determine the relation between the dielectric constant, area of the parallel late and the capacitance of the parallel plate capacitor to solve Part A. The relationship between the capacitance and charge stored in the capacitor are used to find the rank of the capacitors.

Fundamentals

The capacitance of a parallel plate capacitor is given by,

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

Here, $C$ is the capacitance, ${\varepsilon _0}$ is the permittivity of free space, $A$ is the area of cross section of each plate of the capacitor and $d$ is the separation between the plates.

Thus, the dielectric constant of the material between the parallel plates is,

$k = \frac{{Cd}}{{{\varepsilon _0}A}}$

The charge stored in a capacitor is given by,

$Q = CV$

Here, $C$ is the capacitance and $V$ is the potential difference between the plates.

(A)

The dielectric constant of the capacitor is,

$k = \frac{{Cd}}{{{\varepsilon _0}A}}$

Here, the separation between the plates $d$ and the permittivity of free space ${\varepsilon _0}$are constant.

Therefore,

$K = \frac{d}{{{\varepsilon _0}}}$

Here, $K$ is constant.

The equation of dielectric is,

$k = K\frac{C}{A}$

Therefore,

$k\, \propto \,\frac{C}{A}$

Thus, the dielectric constant is directly proportional to$\frac{C}{A}$.

The capacitors are numbered as 1, 2,3,4,5 and 6.

Substitute $4\,{\rm{c}}{{\rm{m}}^2}$ for$A$ and $2\,{\rm{nF}}$ for the capacitance for first capacitor to find the dielectric constant of the first capacitor.

$\begin{array}{c}\\{k_1} = \left( K \right)\left( {\frac{{\left( {2\,{\rm{nF}}} \right)\left( {\frac{{{{10}^{ - 9}}{\rm{ F}}}}{{1\,{\rm{nF}}}}} \right)}}{{\left( {4\,{\rm{c}}{{\rm{m}}^2}} \right)\left( {\frac{{{{10}^{ - 4}}{{\rm{m}}^2}}}{{1\,{\rm{c}}{{\rm{m}}^2}}}} \right)}}} \right)\\\\ = \left( {0.5 \times {{10}^{ - 5}}\,{\rm{F/}}{{\rm{m}}^2}} \right)K\\\end{array}$

Here, ${k_1}$ is the dielectric constant of the first capacitor.

Substitute $1\,{\rm{c}}{{\rm{m}}^2}$ for$A$ and $1\,{\rm{nF}}$ for the capacitance for second capacitor to find the dielectric constant of the second capacitor.

$\begin{array}{c}\\{k_2} = \left( K \right)\left( {\frac{{\left( {1\,{\rm{nF}}} \right)\left( {\frac{{{{10}^{ - 9}}{\rm{ F}}}}{{1\,{\rm{nF}}}}} \right)}}{{\left( {1\,{\rm{c}}{{\rm{m}}^2}} \right)\left( {\frac{{{{10}^{ - 4}}{{\rm{m}}^2}}}{{1\,{\rm{c}}{{\rm{m}}^2}}}} \right)}}} \right)\\\\ = \left( {1 \times {{10}^{ - 5}}\,{\rm{F/}}{{\rm{m}}^2}} \right)\left( K \right)\\\end{array}$

Here, ${k_2}$ is the dielectric constant of the second capacitor.

Substitute $2\,{\rm{c}}{{\rm{m}}^2}$ for$A$ and $8\,{\rm{nF}}$ for the capacitance for third capacitor to find the dielectric constant of the third capacitor.

$\begin{array}{c}\\{k_3} = \left( K \right)\left( {\frac{{\left( {8\,{\rm{nF}}} \right)\left( {\frac{{{{10}^{ - 9}}{\rm{ F}}}}{{1\,{\rm{nF}}}}} \right)}}{{\left( {2\,{\rm{c}}{{\rm{m}}^2}} \right)\left( {\frac{{{{10}^{ - 4}}{{\rm{m}}^2}}}{{1\,{\rm{c}}{{\rm{m}}^2}}}} \right)}}} \right)\\\\ = \left( {4 \times {{10}^{ - 5}}\,{\rm{F/}}{{\rm{m}}^2}} \right)\left( K \right)\\\end{array}$

Here, ${k_3}$ is the dielectric constant of the third capacitor.

Substitute $8\,{\rm{c}}{{\rm{m}}^2}$ for$A$ and $2\,{\rm{nF}}$ for the capacitance for fourth capacitor to find the dielectric constant of the fourth capacitor.

$\begin{array}{c}\\{k_4} = \left( K \right)\left( {\frac{{\left( {2\,{\rm{nF}}} \right)\left( {\frac{{{{10}^{ - 9}}{\rm{ F}}}}{{1\,{\rm{nF}}}}} \right)}}{{\left( {8\,{\rm{c}}{{\rm{m}}^2}} \right)\left( {\frac{{{{10}^{ - 4}}{{\rm{m}}^2}}}{{1\,{\rm{c}}{{\rm{m}}^2}}}} \right)}}} \right)\\\\ = \left( {0.25 \times {{10}^{ - 5}}\,{\rm{F/}}{{\rm{m}}^2}} \right)\left( K \right)\\\end{array}$

Here, ${k_4}$ is the dielectric constant of the fourth capacitor.

Substitute $4\,{\rm{c}}{{\rm{m}}^2}$ for$A$ and $1\,{\rm{nF}}$ for the capacitance for fifth capacitor to find the dielectric constant of the fifth capacitor.

$\begin{array}{c}\\{k_5} = \left( K \right)\left( {\frac{{\left( {1\,{\rm{nF}}} \right)\left( {\frac{{{{10}^{ - 9}}{\rm{ F}}}}{{1\,{\rm{nF}}}}} \right)}}{{\left( {4\,{\rm{c}}{{\rm{m}}^2}} \right)\left( {\frac{{{{10}^{ - 4}}{{\rm{m}}^2}}}{{1\,{\rm{c}}{{\rm{m}}^2}}}} \right)}}} \right)\\\\ = \left( {0.25 \times {{10}^{ - 5}}\,{\rm{F/}}{{\rm{m}}^2}} \right)\left( K \right)\\\end{array}$

Here, ${k_5}$ is the dielectric constant of the fifth capacitor.

Substitute $2\,{\rm{c}}{{\rm{m}}^2}$ for$A$ and $4\,{\rm{nF}}$ for the capacitance for sixth capacitor to find the dielectric constant of the sixth capacitor.

$\begin{array}{c}\\{k_6} = \left( K \right)\left( {\frac{{\left( {4\,{\rm{nF}}} \right)\left( {\frac{{{{10}^{ - 9}}{\rm{ F}}}}{{1\,{\rm{nF}}}}} \right)}}{{\left( {2\,{\rm{c}}{{\rm{m}}^2}} \right)\left( {\frac{{{{10}^{ - 4}}{{\rm{m}}^2}}}{{1\,{\rm{c}}{{\rm{m}}^2}}}} \right)}}} \right)\\\\ = \left( {2 \times {{10}^{ - 5}}\,{\rm{F/}}{{\rm{m}}^2}} \right)\left( K \right)\\\end{array}$

Here, ${k_6}$ is the dielectric constant of the sixth capacitor.

Therefore, the ranking of the dielectric constants is,

${k_3} > {k_6} > {k_2} > {k_1} > \left( {{k_4} = {k_5}} \right)$

The ranking of the capacitors according to the dielectric constant is,

$3 > 6 > 2 > 1 > \left( {4 = 5} \right)$

The charge stored in the capacitor is,

$Q = CV$

For constant potential difference $V$ the charge stored in the capacitor is directly proportional to the capacitance.

$Q \propto V$

The capacitance of the capacitors 1, 2, 3, 4, 5, 6 are ${C_1},\,{C_2},\,{C_3},\,{C_4},\,{C_5}\,{\rm{and}}\,{C_6}$ respectively. The value of ${C_1}$ is $2\,{\rm{nF}}$, ${C_2}$ is $1\,{\rm{nF}}$,${C_3}$ is $8\,{\rm{nF}}$,${C_4}$ is $2\,{\rm{nF}}$,${C_5}$ is $1\,{\rm{nF}}$ and${C_6}$ is $4\,{\rm{nF}}$.

The ranking of the capacitors according to the capacitance is,

${C_3} > {C_6} > \left( {{C_1} = {C_4}} \right) > \left( {{C_2} = {C_5}} \right)$

The ranking of the capacitors with respect to the charge stored is,

$3 > 6 > \left( {1 = 4} \right) > \left( {2 = 5} \right)$

Ans: Part A

The ranking of the capacitors according to the dielectric constant is,

$3 > 6 > 2 > 1 > \left( {4 = 5} \right)$

The ranking of the capacitors with respect to the charge stored is,

$3 > 6 > \left( {1 = 4} \right) > \left( {2 = 5} \right)$