Problem 1:

A negative charge, -Q, is located at the center of a NON- metallic spherical shell of inner radius a, outer radius b. The shell is filled with a uniform distribution of charge of density . The electric field outside the shell (r>b) is

E = +Q/(4 e0r2)r

where R is measured from the center of the shell (r is a unit vector here).

  1. What is the value of E for r < a (i.e. inside the shell)?
  2. What is the value of the total charge distributed within the shell itself?
  3. What is the value of the E field within the shell (i.e. a < r < b)?

Solution:

  1. Clearly, the amount of symmetry in the problem suggests the use of Gauss' law.

    For r < a, we construct a spherical Gaussian surface at some radius r (concentric with the shell). The symmetry of the situation tells us that the electric field is perpendicular to the surface and the same magnitude everywhere on that surface. Then Gauss' law gives

      E · n dA   =   EA  =   4k Qinside

    or

        E = kQinside/r2 r

    since A = 4 r2.

    Thus the only question is: What is Qinside? Well, the only charge inside of the Gaussian surface is the charge -Q; thus we have

        E(r < a)   =  -kQ/r2 r

    (of course this is just the field from a point charge, so we could have written it down immediately...)

  2. It might seem like we don't have enough information to find the total charge in the shell - but we are given the value of the field outside the shell so we can construct a Gaussian surface at r > b (outside the shell, and apply Gauss's law again:

    E A   =   4k Qinside

    where again A = 4 r2. But we are given the value of E, so we can substitute this into the left hand side:

    E A   =   +Q/(4 e0r2) A   =   +kQA/r2

    Substituting for A, and equating the left and right hand sides we get:

    Qinside = +Q

    Now, for this Gaussian surface, what is Qinside? Well, it still includes our charge at the center (-Q) and it now includes the total charge on the shell Qshell. Thus we have:

    Qshell + (-Q)  =   Qinside  =   +Q

    or    Qshell = +2Q .

    Again, one could look at the result for the field at r > b and recognize it as the field from a point charge of +Q; the only way to get that field from a point charge of -Q and a shell of charge is for the shell of charge to have total charge +2Q.

  3. This is somewhat harder (but we have done examples just like this in the problem session). For a < r < b not all of the shell's charge is inside our Gaussian surface; the amount inside depends on r. Since the charge is uniformly distributed in the shell, the fraction of the shell's total charge inside the shell is just the fraction of the shell's total volume that is inside the Gaussian surface. Thus we have

       E A = 4k Qinside

    where

       Qinside = +2Q[(4/3 r3 - 4/3 a3) / (4/3 b3 - 4/3 a3)] - Q

    (the very last term is the charge at the center, which we can't forget!). So we have

    E = k Qinside/r2

    where Qinside is as given above. It can be made to look prettier by cancelling lots of factors of 4/3 if you like...

    Problem 2
    Test 2
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    last updated: March 25 1998