There is a very large horizontal conducting plate in the x−y plane (consider it infinitely large). Its thickness is h. The charge densities on the upper and the lower surfaces are both equal to +σ.
What is the magnitude of the electric field? Express your answer in terms of, if necessary, σ, h and the constant ϵ0.
(b) Consider now a point 3h below the lower surface. What is the direction of the electric field?
(c) Consider now a point h/4 below the top surface.
What is the magnitude of the electric field? Express your answer in terms of, if necessary, σ, h and the constant ϵ0.
a-sigma/epsilon_0
b-sigma/epsilon_0
c-0.
To find the magnitude of the electric field at a point near a large conducting plate, we can use Gauss's law. Gauss's law states that the electric flux through any closed surface is proportional to the total charge enclosed by that surface.
(a) To find the magnitude of the electric field, we can draw a Gaussian surface in the form of a rectangular box that crosses the conducting plate. Assume the side lengths of the box are L1, L2, and L3, and it is centered on the point where we want to find the electric field.
Since the plate is infinitely large, the electric field will be uniform in magnitude and direction within the region of the Gaussian surface. The electric field will point perpendicular to the conducting plate.
The total charge enclosed within the Gaussian surface is zero since charge resides only on the surfaces of the plate, and there is no charge within the plate.
Using Gauss's law, the electric flux through the Gaussian surface is given by:
Flux = Electric field * Area = E * (L1 * L2)
The total charge enclosed by the Gaussian surface is zero, so the electric flux through the surface is also zero. Therefore, we have:
Electric field * (L1 * L2) = 0
Since L1 and L2 are both non-zero, the electric field magnitude E must be zero.
Therefore, the magnitude of the electric field at any point outside the conducting plate is zero.
(b) Consider a point 3h below the lower surface. Since the lower surface has a positive charge density, the electric field will point away from the surface. The direction of the electric field at this point will be upward, perpendicular to the conducting plate.
(c) Consider a point h/4 below the top surface. Since the top surface also has a positive charge density, the electric field will point away from the surface. The point is still above the plate, so the direction of the electric field will also be upward, perpendicular to the conducting plate.
The magnitude of the electric field at this point can be found using the same logic as in part (a). By drawing a Gaussian surface centered on the point and using Gauss's law, we can conclude that the magnitude of the electric field at any point above the conducting plate is zero. Therefore, the magnitude of the electric field at the point h/4 below the top surface is also zero.