Question

Drive an expression for the electric field intensity at the surface of te plate

Answer 1

Answer:

The Gauss law states that electric flux passing through any closed surface is equal to the charge enclosed by that surface divided by permittivity of vacuum.

By symmetry, the magnitude of the electric field will be the same at all points on the curved surface of the cylinder and directed radially outward.

E

and

ds

are along the same direction.

Using this law derive an expression for the electric field due to a uniformly charged infinite plane sheet. Solution : Electric field due to a uniformly charged infinite plane sheet : Suppose a thin non-conducting infinite sheet of uniform surface, charge density σ.

Explanation:

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