Question

1. Below is a hemispherical bowl (basically, half of a hollow sphere) with radius R and uniform charge per unit area: Figure 1: Problem 1 What is the electric field at the center of the hemisphere at the point P? To clarify, imagine we have a sphere instead of half a sphere. The point P corre sponds to the center of this sphere, so a height R above the bottom of the bowl.

Answer 1

Now lets consider Q charge on the sphere and A be the surface area. Charge density sigma = Q/A equation (1) sigma = constant rightarrow so cut it into hemisphere. from Gauss's law phi_sphere = Charge Enclosed/elementof_0 phi_hemisphere = charge enclosed/2 elementof_0 phi_hemisphere = Q/4 elementof_0 equation (2) [charge on hemisphere is Q/2] But we also know that phi = E^vector middot A^vector = EA Note here phi_in = phi_out = EA Lets take phi_out = phi_hemisphere equation (3) from equation (1), (2) and (3) phi_hemisphere = Q/4 elementof_0 = EA E = Q/4A elementof_0 E = sigma/4 elementof_0