Question

Six parallel-plate capacitors of identical plate separation have different plate areas A, different capacitances C, and different dielectrics filling the space between the plates. Part A. Rank the following capacitors on the basis of the dielectric constant of the material between the plates. Rank from largest to smallest.

1. \(A=4 \mathrm{~cm}^{2}  \quad \text {C}=2 \mathrm{nF}\)

2. \(A=1 \mathrm{~cm}^{2} \quad \mathrm{C}=1 \mathrm{nF}\)

3. \(A=2 \mathrm{~cm}^{2}  \quad \mathrm{C}=8 \mathrm{nF}\)

4. \(A=8 \mathrm{~cm}^{2}  \quad \mathrm{C}=2 \mathrm{nF}\)

5. \(A=4 \mathrm{~cm}^{2}  \quad \mathrm{C}=1 \mathrm{nF}\)

6. \(A=2 \mathrm{~cm}^{2}  \quad \mathrm{C}=4 \mathrm{nF}\)

All of the capacitors from Part A are now attached to batteries with the same potential difference. Rank the capacitors on the basis of the charge 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 and the capacitor's charge. 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 relationship between the dielectric constant, area of the parallel plate, and the parallel plate capacitor's capacitance to solve Part A. The relationship between the capacitance and charge stored in the capacitor is used to find the capacitors' rank.

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, \(k=\frac{C d}{\varepsilon_{0} A}\)

The charge stored in a capacitor is given by, \(Q=C V\)

Here, \(C\) is the capacitance, and \(V\) is the potential difference between the plates.

 

(A) The dielectric constant of the capacitor is, \(k=\frac{C d}{\varepsilon_{0} A}\)

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 \(4 \mathrm{~cm}^{2}\) for \(A\) and \(2 \mathrm{nF}\) for the capacitance for first capacitor to find the dielectric constant of the first capacitor.

$$ \begin{aligned} k_{1} &=(K)\left(\frac{(2 \mathrm{nF})\left(\frac{10^{-9} \mathrm{~F}}{1 \mathrm{nF}^{-}}\right)}{\left(4 \mathrm{~cm}^{2}\right)\left(\frac{10^{-4} \mathrm{~m}^{2}}{1 \mathrm{~cm}^{2}}\right)}\right) \\ &=\left(0.5 \times 10^{-5} \mathrm{~F} / \mathrm{m}^{2}\right) K \end{aligned} $$

Here, \(k_{1}\) is the dielectric constant of the first capacitor. Substitute \(1 \mathrm{~cm}^{2}\) for \(A\) and \(1 \mathrm{nF}\) for the capacitance for second capacitor to find the dielectric constant of the second capacitor.

$$ \begin{aligned} k_{2} &=(K)\left(\frac{(1 \mathrm{nF})\left(\frac{10^{-9} \mathrm{~F}}{1 \mathrm{nF}^{-}}\right)}{\left(1 \mathrm{~cm}^{2}\right)\left(\frac{10^{-4} \mathrm{~m}^{2}}{1 \mathrm{~cm}^{2}}\right)}\right) \\ &=\left(1 \times 10^{-5} \mathrm{~F} / \mathrm{m}^{2}\right)(K) \end{aligned} $$

Here, \(k_{2}\) is the dielectric constant of the second capacitor. Substitute \(2 \mathrm{~cm}^{2}\) for \(A\) and \(8 \mathrm{nF}\) for the capacitance for third capacitor to find the dielectric constant of the third capacitor.

$$ \begin{aligned} k_{3} &=(K)\left(\frac{(8 \mathrm{nF})\left(\frac{10^{-9} \mathrm{~F}}{1 \mathrm{nF}^{-}}\right)}{\left(2 \mathrm{~cm}^{2}\right)\left(\frac{10^{-4} \mathrm{~m}^{2}}{1 \mathrm{~cm}^{2}}\right)}\right) \\ &=\left(4 \times 10^{-5} \mathrm{~F} / \mathrm{m}^{2}\right)(K) \end{aligned} $$

Here, \(k_{3}\) is the dielectric constant of the third capacitor. Substitute \(8 \mathrm{~cm}^{2}\) for \(A\) and \(2 \mathrm{nF}\) for the capacitance for fourth capacitor to find the dielectric constant of the fourth capacitor.

$$ k_{4}=(K)\left(\frac{(2 \mathrm{nF})\left(\frac{10^{-9} \mathrm{~F}}{1 \mathrm{nF}^{-}}\right)}{\left(8 \mathrm{~cm}^{2}\right)\left(\frac{10^{-4} \mathrm{~m}^{2}}{1 \mathrm{~cm}^{2}}\right)}\right) $$

\(=\left(0.25 \times 10^{-5} \mathrm{~F} / \mathrm{m}^{2}\right)(K)\)

Here, \(k_{4}\) is the dielectric constant of the fourth capacitor. Substitute \(4 \mathrm{~cm}^{2}\) for \(A\) and \(1 \mathrm{nF}\) for the capacitance for fifth capacitor to find the dielectric constant of the fifth capacitor.

$$ \begin{array}{l} k_{5}=(K)\left(\frac{(1 \mathrm{nF})\left(\frac{10^{-9} \mathrm{~F}}{1 \mathrm{nF}}\right)}{\left(4 \mathrm{~cm}^{2}\right)\left(\frac{10^{-4} \mathrm{~m}^{2}}{1 \mathrm{~cm}^{2}}\right)}\right) \\ =\left(0.25 \times 10^{-5} \mathrm{~F} / \mathrm{m}^{2}\right)(K) \end{array} $$

Here, \(k_{5}\) is the dielectric constant of the fifth capacitor. Substitute \(2 \mathrm{~cm}^{2}\) for \(A\) and \(4 \mathrm{nF}\) for the capacitance for sixth capacitor to find the dielectric constant of the sixth capacitor.

$$ \begin{aligned} k_{6} &=(K)\left(\frac{(4 \mathrm{nF})\left(\frac{10^{-9} \mathrm{~F}}{1 \mathrm{nF}}\right)}{\left(2 \mathrm{~cm}^{2}\right)\left(\frac{10^{-4} \mathrm{~m}^{2}}{1 \mathrm{~cm}^{2}}\right)}\right) \\ &=\left(2 \times 10^{-5} \mathrm{~F} / \mathrm{m}^{2}\right)(K) \end{aligned} $$

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, \(3>6>2>1>(4=5)\)

 

The charge stored in the capacitor is,\( Q=C V \)

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 \(C_{1}, C_{2}, C_{3}, C_{4}, C_{5}\) and \(C_{6}\) respectively. The value of \(C_{1}\) is \(2 \mathrm{nF}, C_{2}\) is \(1 \mathrm{nF}, C_{3}\) is \(8 \mathrm{nF}, C_{4}\) is \(2 \mathrm{nF}, C_{5}\) is \(1 \mathrm{nF}\) and \(C_{6}\) is \(4 \mathrm{nF}\)

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 concerning the charge stored is, \(3>6>(1=4)>(2=5)\)