Question

a non-conducting solid sphere of radius 8.5 centimeters has a uniform charge density. the magnitude of the electric field at 17 centimeters from the spheres center is 2.14×10^3 Newtons per Coulomb.

a. what is the spheres volume charge density?

b. find the magnitude of the electric field at a distance of 5 centimeters from the sphere center

Answer 1

The electric field at a point lying outside the sphere will be due to the entire charge contained by the sphere. However, for a point lying inside the sphere, we will need to apply Gauss's law to determine the electric field due to the charge enclosed within the Gaussian surface.

We will assume that the volume charge density for the given sphere is D C/m^3

a.) Now we have been given the electric field at a distance of, say r, 17 cms. We will use that to determine the unknown value of X

So, total charge = Volume x charge density = (4/3)*pi*8.5^3 x 10^-6 x D

That is, electric field at 17 cms = KQ/r^2 = 8.99 x 10^9 [(4/3)*pi*8.5^3 x 10^-6 x D] / 17*17 x 10^-4

2.14×10^3 = 8.99 x 10^7 [(4/3)*pi*8.5^3 x D] / 17*17

or, (2.14 x 17 x 17 x 3) / 8.99 x 4 x 8.5^3 = D x 10^4

or, D = (2.14 x 17 x 17 x 3) / (8.99 x 4 xpix 8.5^3 x 10^4) = 0.02676 x 10^-4 c/m^3

b.) Here, we assume that a Gaussian surface cocentric with the given sphere and radius of 5 cms.

Now. by Gauss's law, we have: ∫E.ds = charge enclosed /  ε₀

or, E = 0.02676 x 10^-4 x (4/3)* pi * 5^3 x 10^-6 / 4 * pi * 25 * 10^-4 * 8.854 × 10^-12

or E = 0.02676 x 4* pi * 5^3 x 10^6 / 4 * 3 *pi * 25 * 8.854 = 0.005037 x 10^6 N/C = 5.037 x 10^3 N/C

NOTE: For such questions, you need to take the following approach:

a.) Find the unknown, density in this case, using the given electric field.

b.) Once, the charge density is known you can determine the electric field at any point by simple application of Gauss's law.