Since there is spherical symmetry the electric field is normal to the spherical Gaussian surface, and ~ ¡ dA ~ = E dA, while since it is everywhere the same on this surface. The dot product simplifies to E E is a constant on the surface we can pull it out of the integral, and we end up with I q enc E dA = , 0 H Now dA = 4πr2 , where r is the radius of the Gaussian surface, so E=
q enc . 4Ď€ 0 r2
We aren’t done yet, because the charge enclosed depends on the radius of the Gaussian surface. We need to do part (a) again, except this time we don’t want to do the whole volume of the sphere, we only want to go out as far as the Gaussian surface. Then Z q enc = Ď dV, Z r Ď S r = 4Ď€r2 dr, R 0 4Ď€Ď S 1 4 = r , R 4 r4 = Ď€Ď S . R Combine these last two results and E
= = =
Ď€Ď S r4 , 4Ď€ 0 r2 R Ď€Ď S r2 , 4Ď€ 0 R Q r2 . 4Ď€ 0 R4
In the last line we used the results of part (a) to eliminate Ď S from the expression. P27-18 (a) Inside the conductor E = 0, so a Gaussian surface which is embedded in the conductor but containing the hole must have a net enclosed charge of zero. The cavity wall must then have a charge of −3.0 ÂľC. (b) The net charge on the conductor is +10.0 ÂľC; the charge on the outer surface must then be +13.0 ÂľC. P27-19 (a) Inside the shell E = 0, so the net charge inside a Gaussian surface embedded in the shell must be zero, so the inside surface has a charge −Q. (b) Still −Q; the outside has nothing to do with the inside. (c) −(Q + q); see reason (a). (d) Yes.
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