This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 0 in the conductors, in other places, because of the spherical symmetry, it is just the field of a point charge E(r) = ke Q/r2 . Therefore the potential difference V =
O A E(r)dr = ke Q c R dr + r2 a b dr r2 = ke Q 1 1 1 1  +  . c R a b (6) 4. This problem relies on the conservation of energy: mv 2 = q (i  f ) , 2 where i = ke v= 2q (i  f ) m (7) 2Q 2Q  2 a (a + (l  a)2 )1/2 (8) is the electric potential at the starting point of the bullet and f = ke 2Q 2Q  l a (9) the potential at the point the bullet leaves the tube. The first term on the right of (8,9) is the contribution from the positive charge +2Q, while the second is the contribution from two negative charges Q. Denominators are distances from charges to the corresponding points. 5. When brought close to the conducting plane, both positive charges induce negative charge densities in the plane. Each o...
View Full
Document
 Spring '08
 Ingor
 Charge

Click to edit the document details