Consider the isolated charge distribution, A, above, which consists of a solid c

Question

Consider the isolated charge distribution, A, above, which consists of a solid conducting sphere of radius r=a and a charge +q surrounded by a solid neutral conducting shell of inner radius r=b and outer radius r=c. in terms of given quantities and appropriate constants derive:

a. E(r<a)

bE(a<r<b)

c. E(b<r<c)

d E(r>c)

a. E(r<a)

bE(a<r<b)

c. E(b<r<c)

d E(r>c)

Explanation / Answer

There can be no field within either conductor (or current would flow) The field at b > r > a is E = K q / r^2 The charge +q at a induces -q at r = b This leaves +q on the outer sphere at r = c The field for r > c is E = k q / r^2 If you add -q to the surface of the outer sphere then there is no net charge on the outer surface and E = 0 for r >c For r

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