THE PARALLEL PLATE CAPACITOR

Fig.1: A commercial capacitor is labeledwith the value of its capacitance.

A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance (Fig. 2). We first take the intervening medium between the plates to be vacuum. The effect of a dielectric medium between the plates is discussed in the next section. Let A be the area of each plate and d the separation between them. The two plates have charges Q and –Q. Since d is much smaller than the linear dimension of the plates ($d^2<< A$), we can use the result on electric field by an infinite plane sheet of uniform surface charge density. 

Fig.2

Plate 1 has surface charge density $\sigma = Q/A$ and plate 2 has a surface charge density $-\sigma$. the electric field in different regions is:

Outer region I (region above the plate 1),

$E=\frac{\sigma}{2\epsilon}-\frac{\sigma}{2\epsilon}=0$                 (1)

Outer region II (region below the plate 2),

$E=\frac{\sigma}{2\epsilon}-\frac{\sigma}{2\epsilon}=0$                 (2)

In the inner region between the plates 1 and 2, the electric fields due to the two charged plates add up, giving

$E=\frac{\sigma}{2\epsilon}+\frac{\sigma}{2\epsilon}=\frac{\sigma}{\epsilon}$

$E=\frac{Q}{\epsilon A}$                               (3)

The direction of electric field is from the positive to the negative plate.

Thus, the electric field is localised between the two plates and is uniform throughout. For plates with finite area, this will not be true near the outer boundaries of the plates. The field lines bend outward at the edges — an effect called ‘fringing of the field’. By the same token, $\sigma$ will not be strictly uniform on the entire plate. [E and σ are related by $E=\frac{\sigma}{\epsilon_0}$.] However, for $d^2 << A$, these effects can be ignored in the regions sufficiently far from the edges, and the field there is given by Eq. (3). Now for uniform electric field, potential difference is simply the electric field times the distance between the plates, that is, 

$V=Ed=\frac{Qd}{\epsilon_0 A}$                                    (4)

The capacitance C of the parallel plate capacitor is then

$C=\frac{Q}{V}=\frac{\epsilon_0 A}{d}$                       (5)

which, as expected, depends only on the geometry of the system. For typical values like $A = 1 \ m^2$ , d = 1 mm, we get

$C=\frac{8.85 \times 10^{-12} \ C^2.N^{-1}.m^{-2} \times 1 \ m^2}{10^{-3} \ m}=8.85 \times 10^{-9} \ F$

(You can check that if $1 \ F= 1 \ C.V^{–1} = 1 \ C (N.C^{–1}m)^{–1} = 1 \ C^2.N^{–1}.m^{–1}$.) This shows that 1F is too big a unit in practice, as remarked earlier. Another way of seeing the ‘bigness’ of 1 F is to calculate the area of the plates needed to have $C = 1 \ F$ for a separation of, say 1 cm:

$A=\frac{Cd}{\epsilon_0}=\frac{1 \ F \times 10^{-2} \ m}{8.85 \times 10^{-12} \ C^2.N^{-1}.m^{-2}}=10^9 \ m^2$

which is a plate about 30 km in length and breadth!

Example 1

The plates of a parallel-plate capacitor in vacuum are apart and in area 5.00 mm apart and 2.00 $m^2$ in area. A 10.0 kV potential difference is applied across the capacitor. Compute (a) the capacitance; (b) the charge on each plate; and (c) the magnitude of the electric field between the plates.

Solution

We are given the plate area A, the plate spacing d and the potential difference $V=1.00 \times 10^4 \ V$ for this parallel-plate capacitor. Our target variables are the capacitance C, the charge on each plate, and the charge Q on each plate, and electric-field magnitude E. We use Eq. (4) to calculate C and then use Eq. (3) and $V$ to find Q. We use $E=Q/\epsilon_0A$ to find E.

(a) From Eq.(1)

$C=\frac{\epsilon_0 A}{d}=\frac{(8.85 \times 10^{-12} \ C^2.N^{-1}.m^{-2})(2.00 \ m^2)}{5.00 \times 10^{-3} \ m}$

$C=3.54 \times 10^{-9} \ F=3.54 \ nF$

(b) The charge on the capacitor is

$Q=CV=(3.54 \times 10^{-9} \ F)(1.00 \times 10^4 \ V)$

$Q=3.54 \times 10^{-5} \ C=35.4 \mu C$

The plate at higher potential has charge $+35.4 \ \mu C$, and the other plate has charge $-35.4 \ \mu C$.

(c) The electric-field magnitude is

$E=\frac{\sigma}{\epsilon_0}=\frac{Q}{\epsilon_0 A}$

$E=\frac{3.54 \times 10^{-5} \ C}{(8.85 \times 10^{-12} \ C^2.N^{-1}.m^{-2})(2.00 \ m^2)}$

$E=2.00 \times 10^6 \ N/C$

We can also find E by recalling that the electric field is equal in magnitude to the potential gradient. The field between the plates is uniform, so

$E=\frac{V}{d}=\frac{1.00 \times 10^4 \ V}{5.00 \times 10^{-3} \ m}=2.00 \times 10^6 \ V/m$

Share :