Question: A solid conducting sphere of radius R is cut through its midplane into two hemispheres. The two halves are then brought back together so that

 

A solid conducting sphere of radius R is cut through its midplane into two hemispheres. The two halves are then brought back together so that the two halves are electrically isolated by a narrow gap of thickness S << R. A charge q is then placed on the upper hemisphere, while the tower hemisphere remains uncharged. In this problem, you may ignore edge effects which occur at distances of order S from the equator where the two spheres meet.

a. What is the charge per unit area σ 1 (r) on the planar surface of the upper hemisphere, as a function of distance r from the center of the sphere?

b. What is the charge per unit area σ 2 (r) on the planar surface of the lower hemisphere as a function of the distance r from the center of the sphere?

c. What is the charge per unit area σ 3 (θ) on the outer surface of the upper hemisphere, where θ is the polar angle of a point on the surface, with the north pole at θ = 0 and the equator at θ = π/2?

d. What is the charge per unit area σ 4 (θ) on the outer surface of the lower hemisphere. where θ on the lower hemisphere ranges from π/2 (equator) to π at the south pole?

e. What is the electric field E(r) in the gap between the hemispheres?

f. What is the electric potential difference V between the hemispheres?

 

Step by Step Solution

3.36 Rating (146 Votes )

There are 3 Steps involved in it

1 Expert Approved Answer
Step: 1 Unlock

This problem involves analyzing charge distribution and electric fieldspotentials related to a conducting sphere that is divided into two hemispheres with a small insulating gap We will tackle the pro... View full answer

blur-text-image
Question Has Been Solved by an Expert!

Get step-by-step solutions from verified subject matter experts

Step: 2 Unlock
Step: 3 Unlock

Students Have Also Explored These Related Physics Questions!

Related Book