# Homework Help: Gaussian Surface and Charge Density

1. Jul 30, 2013

### whitehorsey

1. A solid dielectric sphere of radius 10 cm has an electric charge uniformly distributed throughout its volume. The electric field at 5 cm from the center of the sphere is 8.6 x 10^4 N/C, pointing radially outward. Calculate the magnitude and direction of the electric field at a point 15 cm from the center of the sphere.

2. Gaussian Law

3. I solved it like this:

E(4∏r2) = Q(r3/a3)/ ε0
E = Qr/4∏ε0a3
Q = 1.91 * 10^-7C
E = kQ/r2
= 76.4kN/C radically outward.

My question is what is the difference between doing it the above way versus with charge density:
Finding r < a (inside the sphere)
Equation: E(4∏r2) = 4/3(∏r3)ρ/ ε0
ρ = 4.5666 *10^-5

Find r > a (outside the sphere)
E(4∏r2) = 4/3(∏a3)ρ/ ε0
E = 76.4 kN/C radically outward

2. Jul 30, 2013

### Dick

None really. If you use ρ=Q/((4/3)*pi*a^3) (charge/volume) and substitute into your second equation for E(4*pi*r^2) you get the same formula as you used in the first part. The only difference is whether you choose to solve for Q first or ρ first.

Last edited: Jul 30, 2013
3. Jul 31, 2013

Thank You!!