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According to Equation 18.4, the electric field inside a parallel plate capacitor, and away from the edges, is constant and has a magnitude of E =s/e0, where s is the charge density (the charge per unit area) on a plate. Use Gauss’ law to prove this result.
Reasoning
Figure 18.36a shows the electric field inside a parallel plate capacitor. Because the positive and negative charges are distributed uniformly over the surfaces of the plates, symmetry requires that the electric field be perpendicular to the plates. We will take advantage of this symmetry by choosing our Gaussian surface to be a small cylinder whose axis is perpendicular to the plates (see part b of the figure). With this choice, we will be able to evaluate the electric flux and then, with the aid of Gauss’ law, determine E.
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(a) A side view of a parallel plate capacitor, showing some of the electric field lines. (b) The Gaussian surface is a cylinder oriented so its axis is perpendicular to the positive plate and its left end is inside the plate. |
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Solution
Figure 18.36b shows that we have placed our Gaussian cylinder so that its left end is inside the positive metal plate, and the right end is in the space between the plates. To determine the electric flux through this Gaussian surface, we evaluate the flux through each of the three parts—labeled 1, 2, and 3 in the drawing—that make up the total surface of the cylinder and then add up the fluxes.
Surface 1—the flat left end of the cylinder—is embedded inside the positive metal plate. As discussed in Section 18.8, the electric field is zero everywhere inside a conductor that is in equilibrium under electrostatic conditions. Since E=0 N/C, the electric flux through this surface is also zero:
Surface 2—the curved wall of the cylinder—is everywhere parallel to the electric field between the plates, so that cos f=cos 90°=0. Therefore, the electric flux through this surface is also zero:
Surface 3—the flat right end of the cylinder—is perpendicular to the electric field between the plates, so cos f=cos 0°=1. The electric field is constant over this surface, so E can be taken outside the summation in Equation 18.6. Noting that SDA=A is the area of surface 3, we find that the electric flux through this surface is
The electric flux through the entire Gaussian cylinder is the sum of the three fluxes determined above:
According to Gauss’ law, we set the electric flux equal to Q/e0, where Q is the net charge inside the Gaussian cylinder: EA=Q/e0. But Q/A is the charge per unit area, s, on the plate. Therefore, we arrive at the value of the electric field inside a parallel plate capacitor:  . Notice that the distance of the right end of the Gaussian cylinder from the positive plate does not appear in this result, indicating that the electric field has the same value everywhere between the plates.
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