This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Homework #4 1. A metal sphere of radius R, carrying charge q, is surrounded by a thick concentric metal shell (inner radius a, outer radius b). The shell carries no net charge. (a) Find the surface charge density at R, at a, and at b. (b) Find the potential at the center, using infinity as the reference point. (c) Now the outer surface is touched to a ground wire, which lowers its potential to zero (same as at infinity). How do your answers to (a) and (b) change? Solution: (a) The surface charge density at R is obvious ) 4 /( 2 R q R . Since the electric field is zero inside the metal, thus if we choose a sphere between a and b and apply the Gausss law to it, we immediately realize that the total charge at a has to be q. Thus the surface charge density at a is ) 4 /( 2 a q a . Since the total charge in the shell has to be zero, the total charge at the outer surface is +q. Then the charge density at b is ) 4 /( 2 b q b . (b) Applying the Gausss law to a sphere in different regios , it is easy to get the electric field in different regions. E for R r . r r q E 2 4 for a r R . E for b r a . r r q E 2 4 for b r . Then potential at the center is (taking infinity as the reference) b q a q R q dr r q dr r q Edr Edr Edr r d E V b a R b a R 2 2 4 4 4 4 4 (c) If the outer surface of the shell is connected to the ground, the +q on the outer surface will go away. Then the surface charge density at b will be zero while the density at R and at a remain unchanged. The electric field for r>b will also become zero. Then the potential at the center of the sphere will be a q R q dr r q Edr Edr r d E V a R a R 2 4 4 4 . 2. A metal sphere of radius R carries a total charge Q. Whats the force of repulsion between the northern and southern hemisphere? Solution: The surface charge density is 2 4 R Q . Then the force per unit area is n R Q n f 4 2 2 2 32 2 1 . Another way is that the electric field at the surface is n R Q E 2 4 . Then the pressure on the surface is 4 2 2 2 32 2 R Q E P , which is the force per unit area along the surface normal direction....
View Full Document
This note was uploaded on 02/06/2010 for the course PHYSICS 11 taught by Professor Qiu during the Fall '09 term at University of California, Berkeley.
- Fall '09