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...
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- Spring '08
- Charge, Electric charge, Positive charge