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Unformatted text preview: PHYS1002 TUTORIAL 2 SOLUTIONS Semester 2, 2007 1. ELECTRIC POTENTIAL AND FIELD The charges will distribute themselves uniformly on the shells, so the problem has spherical symmetry. The electric field will therefore point in the radial direction (if it is nonzero) and its magnitude can only depend on the distance to the center of the shells. We can therefore first use Gausss law to find the electric field, and then integrate the electric field to find the potential. To find the electric field outside all the shells, i.e. for r > c, we use a Gaussian surface of spherical shape with radius r > c. Since the total charge inside this surface is 0, the net electric flux out of the surface must be 0. This flux can be written EA where 2 4 A r = is the area of the Gaussian surface (this is because E, by the symmetry noted above, must be everywhere perpendicular to this surface and have a constant magnitude on the surface). Therefore we find that E = 0 for r > c. To find the electric field for b < r < c we use a Gaussian surface with radius b < r < c. The argument is exactly the same as above, except that now the total charge inside the surface is +Q. Therefore, solving for E we find that 2 / E kQ r = with 1/(4 ) k = . Since Q > 0, E is positive, i.e. the field points radially outwards. Finally, we use the same procedure to find the field for a < r < b. Because now there is again no charge inside the Gaussian surface with radius a < r < b, the field in this region is zero. Next, we find the potential on the shells. As usual, the potential will be taken to be zero at infinite distance from the shells, ( ) V = . Since the field is zero for r > c, we have ( ) ( ) ( ) 0. c V c V E dr V c = = = a a Furthermore, 2 2 1 1 1 ( ) ( ) ( ) b b b b c c c c kQ dr V b V c E dr r rdr kQ kQ kQ r r r b c = = = = = a a 1 1 ( ) 0. V b kQ b c = > Finally, because the field is zero for a < r < b, we have 1 1 ( ) ( ) ( ) ( ) ....
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 Three '11
 TarasPlank
 Charge, Electric Potential, Magnetism

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