Parallel Plate Capacitor

Section 32.6 Parallel Plate Capacitor

A parallel plate capacitor consists of two large flat metal plates facing each other as shown in Figure 32.37.

The capacitance depends on the area \(A\) of the plates, their separation \(d\text{,}\) and dielectric constant \(\epsilon_r\) of the meterial between the plates.

\begin{equation} C = \frac{\epsilon_0 \epsilon_r\,A}{d},\tag{32.12} \end{equation}

where \(\epsilon_0\) is the permittivity of vacuum.

Derivation of Capacitance Formula for a Parallel Plate Capacitor

Strategy: To deduce the formula given in (32.12), we find the potential difference \(V\) when plates are charged by \(\pm Q\) and then get capacitance from \(V/Q\text{.}\)

Assuming plates to be infinitely large with charge density \(\sigma = Q/A\text{,}\) the electric field in the space between the plates will be constant and directed from positive plate to negativbe plat. If there is vacuum between the plates, electric field wil lhave the following magnitude,

\begin{equation*} E_0 = \frac{Q/A}{\epsilon_0}, \end{equation*}

and if there is a material of dielectric constant, \(\epsilon_r\text{,}\) between the plates, electric field will be

\begin{equation} E = \frac{Q/A}{\epsilon_0\epsilon_r}.\tag{32.13} \end{equation}

Since dieclectric of vacuum is \(\epsilon_{r,\text{vacuum}} = 1\text{,}\) we will just work with Eq. (32.13). With electric field between the plates being constant, the potential difference \(\Delta \phi\) or \(V\) will simply equal the product of the field and the separation of the plates.

\begin{equation*} V= Ed = \frac{Q d}{\epsilon_0\epsilon_r\, A}. \end{equation*}

Now, fundamental capacitor equation tells us that \(Q=CV\text{,}\) i.e., \(C = Q/V\text{.}\) Therefore, capacitance is

\begin{equation} C = \frac{\epsilon_0 \epsilon_r\,A}{d}.\tag{32.14} \end{equation}

This shows that the capacitance of a capacitor depends on the geometrical features of the plates, such as surface area over which charges can spread, the separation between the plates, and the electric permittivity (\(\epsilon_0 \epsilon_r\)) of the medium between the plates.

Subsection 32.6.1 Capacitor with a Dielectric between Plates

Notice the capacitance formula for the parallel plate capacitor with a dielectric between the plates we found in Eq. (32.14). Let’s write the capacitance for the case when we have vacuum between the plates as \(C_0\text{.}\) Setting \(\epsilon_r=1\) in Eq. (32.14) we get the following for \(C_0\text{.}\)

\begin{equation} C_0 = \frac{\epsilon_0\,A}{d}.\tag{32.15} \end{equation}

Using this, we can write the capacitance of the capacitor when we fill the gap by a dielectric material, such as mica or glass, of dielectric constant \(\epsilon_r\text{.}\)

\begin{equation} C = \epsilon_r\, C_0.\tag{32.16} \end{equation}

Since dielectric constant is always greater than \(1\text{,}\) i.e., \(\epsilon_r \gt 1\text{,}\) stuffing the space between the plates with a dielectric will always increase the capacitance. As a matter of fact, this is a very effective way of increasing the capacitance. For instane, if you fill the space between the plates with mica, a commonly used material for this purpose, the capacitance will increase \(7\)-fold since \(\epsilon_{r,\text{mica}} = 7\text{.}\)

Usually mica, cardboard, or paper is put between the plates of a capacitor to increase the capacitance.

If you look at the formula for electric field between the plates when you have dielectric, you notice that with regard to the electric field in vacuum, the electric field with dielectric is reduced by the same factor.

\begin{equation*} E = \frac{E_0}{\epsilon_r}. \end{equation*}

With the reduced electric field between the plates the potential difference between the plates would be less. Therefore, you can put larger amount of charge on the plates for the same potential difference.

Exercises 32.6.2 Exercises

1. Capacitance of Parallel Plate Capacitor as Area and Separation is Varied.

A parallel plate capacitor is constructed from two aluminum foil sheets, each of dimensions \(20 \text{ cm} \times 20 \text{ cm}\text{.}\) The plates are separated by \(2 \text{ mm}\) with nothing between the plates.

(a) Evaluate the capacitance of the capacitor.

(b) What will be the capacitance if the distance between the sheets is increased to \(4 \text{ mm}\text{?}\)

(c) What will be the capacitance if the foils are cut into half making the dimensions \(20 \text{ cm} \times 10 \text{ cm}\) each, but kept at a separation of \(2 \text{ mm}\text{?}\)

Hint.

Use \(C = \epsilon_0 A /d\text{.}\)

Answer.

(a) \(177 \text{ pF}\text{,}\) (b) \(88.5 \text{ pF}\text{,}\) (c) \(88.5 \text{ pF}\text{.}\)

Solution.

(a) The capacitance of a parallel plate capacitor is given by

\begin{align*} C \amp = \frac{\epsilon_0\:A}{d}\\ \amp = \frac{8.85\times 10^{-12}\: \text{F/m}\times (0.2\:\text{m} \times 0.2\:\text{m} )}{0.002\:\text{m}} \\ \amp = 177\:\text{pF}. \end{align*}

(b) If the distance between the plates, \(d\text{,}\) goes up two-fold, then \(C\) would go down by a factor of 2.

\begin{equation*} C = \frac{177\:\text{pF}}{2} = 88.5\:\text{pF} \end{equation*}

(c) The capacitance varies directly as the area of the plates. Since the area became half as much, the capacitance will become half also.

\begin{equation*} C = \frac{177\:\text{pF}}{2} = 88.5\:\text{pF} \end{equation*}

2. Capacitor with Mica between Plates.

The space between two copper plates of dimensions \(20\text{ cm} \times 30\text{ cm}\) separated by \(4\text{ mm}\) is filled with mica.

(a) Find the capacitance.

(b) What is the percent by which the capacitance increased due to filling the space with mica?

Data: \(\epsilon_{r,\text{ mica}} = 7\text{.}\)

Hint.

Use formula.

Answer.

(a) \(0.93\text{ nF}\text{,}\) (b) 600%.

Solution 1. (a)

(a) Using the formula for the capacitance of a parallel plate capacitor with a dielctric we get

\begin{align*} C \amp = K\:\frac{\epsilon_0\:A}{d}\\ \amp = 7\times \frac{8.85\times 10^{-12}\:\textrm{F/m}\times 0.06\:\textrm{m}^2}{0.004\:\textrm{m}} = 0.93\: \textrm{nF}. \end{align*}

Solution 2. (b)

(b) Since the capacitance increases by a factor of \(\epsilon_r\text{,}\) which is 7.

\begin{equation*} \frac{C-C_0}{C_0} = \epsilon_r - 1 = 6, \end{equation*}

which is 600%.

3. Polarization of a Dielectric Between Charged Plates.

Two pieces of aluminum foil of dimensions \(25\text{ cm} \times25\text{ cm}\) is separated by a piece of paper of thickness \(0.2\) mm. (a) Find the capacitance. (b) If the foils are oppositely charged with \(\pm 50\ \mu\text{C}\text{,}\) what is the polarization of paper?

Data: \(\epsilon_{r,\text{ paper}} = 3.7 \text{.}\)

Hint.

(a) Use Capacitance formula, (b) Polarization in a linear dielectric is \(P = \epsilon_0\: (\epsilon_r-1)\: E_0/\epsilon_r\text{,}\) where \(E_0\) is the electric field without the dielectric.

Answer.

(a) \(10.2\text{ nF}\text{,}\) (b) \(580\ \mu\text{C/m}^2\text{.}\)

Solution 1. (a)

(a) Using the formula for the capacitance of a parallel plate capacitor filled with a dielectric we get

\begin{align*} C \amp = \epsilon_r\:\frac{\epsilon_0\:A}{d} \\ \amp = 3.7\times \frac{8.85\times 10^{-12}\:\textrm{F/m}\times 0.0625\:\textrm{m}^2}{0.0002\:\textrm{m}} = 10.2\: \textrm{nF}. \end{align*}

Solution 2. (b)

(b) Polarization depends on the electric field in the dielectric, i.e., paper here, which is related to the electric field in vacuum through the dielectric constant.

\begin{align*} P \amp = \epsilon_0\: (\epsilon_r-1)\: E\\ \amp = \epsilon_0\: (\epsilon_r-1)\: \frac{E_0}{\epsilon_r}\\ \amp = \frac{\epsilon_0\: (\epsilon_r-1)}{\epsilon_r}\:\frac{Q/A}{\epsilon_0},\\ \amp = \frac{ \epsilon_r-1}{\epsilon_r}\:\frac{Q}{A} \end{align*}

Putting the numbers in the expression we find

\begin{align*} P \amp = \frac{3.7-1}{3.7}\:\frac{50\times 10^{-6}\:\textrm{C}}{0.25\times0.25\:\textrm{m}^2}\\ \amp = 5.8\times 10^{-4} \text{C/m}^2. \end{align*}

Prev Top Next