Field Between Oppositely Charged Parallel Conducting Plates

Two large plane parallel conducting plates are given charges of equal magnitude and opposite sign; the surface charge densities are $+\sigma$ and $-\sigma$. Find the electric field in the region between the plates.

Fig.1a shows the field. Because opposite charges attract, most of the charge accumulates at the opposing faces of the plates. A small amount of charge resides on the outer surfaces of the plates, and there is some spreading or “fringing” of the field at the edges. But if the plates are very large in comparison to the distance between them, the amount of charge on the outer surfaces is negligibly small, and the fringing can be neglected except near the edges. In this case we can assume that the field is uniform in the interior region between the plates, as in Fig. 1b, and that the charges are distributed uniformly over the opposing surfaces. To exploit this symmetry, we can use the shaded Gaussian surfaces $S_1$, $S_2$, $S_3$ and $S_4$. These surfaces are cylinders with flat ends of area ; one end of each surface lies A within a plate.

Fig.1: Electric field between oppositely charged parallel plates.

The left-hand end of surface $S_1$ is within the positive plate 1. Since the field is zero within the volume of any solid conductor under electrostatic conditions, there is no electric flux through this end. The electric field between the plates is perpendicular to the right-hand end, so on that end, $E_{perp}$ is equal to and the flux is EA; this is positive, since E is directed out of the Gaussian surface. There is no flux through the side walls of the cylinder, since these walls are parallel to E. So the total flux integral in Gauss’s law is EA. The net charge enclosed by the cylinder is $\sigma A$ so Eq. (22.8) yields $EA=\sigma A/\epsilon_0$ we then have

$E=\frac{\sigma}{\epsilon_0}$ (field between oppositely charged conducting plates)

The field is uniform and perpendicular to the plates, and its magnitude is independent of the distance from either plate. The Gaussian surface $S_4$ yields the same result. Surfaces $S_2$ and $S_3$ yield E = 0 to the left of plate 1 and to the right of plate 2, respectively. We leave these calculations to you (see Exercise 22.29).

We obtained the same results in Example 21.11 by using the principle of superposition of electric fields. The fields due to the two sheets of charge (one on each plate) are $\mathbf{E_1}$ and $\mathbf{E_2}$ from Example 22.7, both of these have magnitude $\sigma/2\epsilon_0$. The total electric field at any point is the vector sum $\mathbf{E=E_1+E_2}$. At points and in Fig. 22.21b, $\mathbf{E_1}$ and $\mathbf{E_2}$ point in opposite directions, and their sum is zero. At point b, $\mathbf{E_1}$ and $\mathbf{E_2}$ are in the same direction; their sum has magnitude just as we found above using Gauss’s law. $E=\sigma/\epsilon_0$,

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