Question

Concentric conducting spherical shells carry charges
Q and –Q, respectively (see below). The inner shell has
negligible thickness. Determine the electric field for (a)
r < a; (b) a < r < b; (c) b < r < c; and (d) r > c.

Please explained as much as you can, thank you.

Answer 1

The inner shell has charge Q and outer shell has charge -Q

This is a case where charge distribution has spherical symmetry , and hence E field is radial. In such cases E can be determined by using Gauss's law , which states that

, where the integral on left hand side(LHS), is evaluated over any closed surface (called Gausian Surface) , and it gives electric flux over that surface,and is the net electric charge enclosed by the chosen Gausian surface and is the permittivity of free space( since there is no dielctric anywhere in the regions under consideration).

Note : E.ds above is dot product of vector E and area element vector ds = = Eds , for , ( since in this case E vector and area vector are parallel)

So in case (a) and (d) obviously Electic field flux = 0 since q ,enclosed by gausian surfaces (spherical) imagined at these radiai enclose no charges [ in (a) no charge at all and in (d), net charge =+Q+ (-Q) =0 ]

So in (a) and (d),

Flux = 0 and hence E=0 , since surface are of Gausian surface is "not equal" to 0

What about (c) ? The field inside any metal is 0 , since charge resides on the surface.

There is another way of looking at this: Since inner sphere has charge +Q a charge of -Q is induced on inner surface of outer shell. As a result the flux in any Gausian surface we draw in region r>a and r<b, that is inside the outer shell,

net charge is 0

Hence in (c) , E=0

In case of region in (b) , the charge enclosed is Q and hence ,if we take a spherical Gausian surface at radius r (  a < r < b ),

then flux =

The field is similar in value to that due to a point charge at r =0