A solid insulating sphere of radius a = 3.0 cm, has a uniform volume charge dens

Question

A solid insulating sphere of radius a = 3.0 cm, has a uniform volume charge density of rho = 95 nC/cm^3 distributed throughout the sphere. Surrounding and concentric with this sphere is a conducting shell with inner radius b = 8.0 cm, and outer radius c = 9.0 cm. (See diagram). The conducting shell has a total charge of Q = - 6.5 muC placed on it. a) What is the total charge on the insulating sphere? b) How much charge is on the inner surface (surface with radius b) of the conducting shell? c) What is the surface charge density on the inner surface of the conducting shell? d) How much charge is on the outer surface (surface with radius c) of the conducting shell? e) What is the electric field at a point r from the center of the insulating sphere, with r > c?

Explanation / Answer

For solving such problems you need to keep the following into consideration:

1.) For an insulating solid with a volume charge distribution, the charged do not redistribute themselves and stay at the points inside the solid as well.

2.) For the conducting shell, as the outer one in the diagram above, the charges will redistribute themselves onto the surface as it is not possible to have an electric field inside the body of a conducting solid, or else a continuous current would start flowing which is not possible,

3.) Also, as the net electric field inside the body of the outer solid must be zero, the net charge enclosed by a Gaussian surface with radius such that b < r < c, must also be zero. [Gauss' law states that the electric flux through a surface is equal to the charge enclosed divided epsilon naught. So for field to remain zero, the net charge included must also be zero]

We will make use of the above points to answer the questions given as follows:

a.) The volume charge density of the solid is 95 nC/m^3

Therefore, the net charge on the insulating sphere would be Volume x density

or, Charge = 95 x 4 / 3 x 3.14 x 27 = 10738.8 nC = 10.7388 C is the required value.

b.) Now, as mentioned in the point 3. above, the net charge enclosed in the Gaussian surface inside the outer shell should be zero. That would mean that the inner surface of the shell must have a charge equal in magnitude and opposite in sign as that of the solid sphere inside.

Therefore, the charge on the inner surface of the shell would be: -10.7388 C

c.) The surface charge density on the inner surface would be equal to the charge contained on unit area which can be obtained by the dividing the charge contained on the surface by its area.

Hence, the surface charge density = -10.7388 / (4 x 3.14 x 64) = -0.013359375 = -13.3594 nC/cm2

d.) Now, as the inner suface of the shell has a charge of -10.7388 C and the net charge on the shell is -6.5 C

The charge on the outer would be given as: Total charge - Charge on inner surface

or Charge on the outer surface = -6.5 + 10.7388 = 4.2388 C

e.) For electric field at a point r>c let us assume a spherical Gaussian surface with radius r and concentric with the given spheres. By Gauss' law we have:

E(4r2) = 10.7388 - 6.5 / 0 = 4.2388 / 0

or, E = 4.2388 / 4r20 is the required expression for the electric field for a point r > c

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