There is a very large horizontal conducting plate in the plane (consider it infinitely large). Its thickness is . The charge densities on the upper and the lower surfaces are both equal to .
The goal of this problem is to find the electric field (magnitude and direction) at different points.
(a) Consider a point above the upper surface. What is the direction of the electric field?
What is the magnitude of the electric field? Express your answer in terms of, if necessary, , and the constant .
unanswered
(b) Consider now a point below the lower surface. What is the direction of the electric field?
What is the magnitude of the electric field? Express your answer in terms of, if necessary, , and the constant .
unanswered
(c) Consider now a point below the top surface.
What is the magnitude of the electric field? Express your answer in terms of, if necessary, , and the constant .
unanswered
Please do not post any more questions until you show your work for the previous questions.
To find the electric field at different points above and below the conducting plate, we can use Gauss's law, which relates the electric field to the charge enclosed by a surface.
(a) To find the direction of the electric field above the upper surface, we need to consider the symmetry of the problem. Since the charge densities on both the upper and lower surfaces are equal, the plate is electrically neutral, and the electric field should be directed perpendicularly away from the plate.
To find the magnitude of the electric field at point P, we can imagine drawing a Gaussian surface in the shape of a pillbox, with its cylindrical part directly above the point P. The electric field passing through this surface will only come from the upper surface of the plate since the charge on the lower surface is inside the Gaussian surface.
Using Gauss's law, we can write:
Electric flux = Electric field x Area = (Charge enclosed)/(Epsilon naught)
Since the plate is infinitely large, the area of the upper surface of the plate is also infinitely large, and the electric flux through this surface will be constant. Therefore, the electric field at point P above the upper surface of the plate is also constant.
Assuming the charge density on both surfaces is denoted by sigma, the charge enclosed by the Gaussian surface is sigma x Area = sigma x A, where A is the area of the cylindrical part of the Gaussian surface.
Using Gauss's law, we can now solve for the electric field:
Electric field x 2πrL = (sigma x A) / (Epsilon naught)
Simplifying, we get:
Electric field = (sigma / (2 x epsilon naught)) x (1 / r)
So, the magnitude of the electric field at point P above the upper surface is (sigma / (2 x epsilon naught)) x (1 / r), while the direction is perpendicular away from the surface.
(b) Similarly, to find the direction of the electric field below the lower surface, we consider that the plate is electrically neutral, and the electric field should be directed perpendicularly towards the plate. The magnitude of the electric field can be found using the same reasoning as in part (a), but this time the Gauss's surface will enclose the charge on the lower surface.
The magnitude of the electric field at point Q below the lower surface is also (sigma / (2 x epsilon naught)) x (1 / r), and the direction is perpendicular towards the surface.
(c) For a point R below the top surface, the contribution from the upper surface is cancelled by the contribution from the lower surface because the plate is electrically neutral. Therefore, the magnitude of the electric field at point R is zero, and the direction is undefined as there is no net electric field.
In summary:
(a) Electric field magnitude above the upper surface: (sigma / (2 x epsilon naught)) x (1 / r) directed away from the surface.
(b) Electric field magnitude below the lower surface: (sigma / (2 x epsilon naught)) x (1 / r) directed towards the surface.
(c) Electric field magnitude below the top surface: 0, direction is undefined.