# Homework Help: Find the electric field strength inside a solid sphere.

1. Oct 1, 2008

### _F_

1. The problem statement, all variables and given/known data
A solid sphere 25cm in radius carries 14uC, distributed uniformly throughout its volume. Find the electric field strength (a.) 15cm, (b.) 25cm, (c.) 50cm from the sphere's center.

R = .25m
Q = 14 * 10^-6 C

2. Relevant equations
p = q/(volume)
E = (kq)/r^2

3. The attempt at a solution
a.) I have absolutely no idea where to even begin here. This is what I really need help with. I'm not sure if the answer relates to the other two distances, so i'll show how I got those. For some reason, E = (kq)/r^2 fails here.

b.) E = (kq)/r^2 = (9.0*10^9 Nm^2/C^2)(14*10^-6 C)/(.25m)^2 = 2.02 * 10^-6 N/C

c.) E = (kq)/r^2 = (9.0*10^9 Nm^2/C^2)(14*10^-6 C)/(.50m)^2 = 504 * 10^3 N/C

2. Oct 1, 2008

### gabbagabbahey

The field inside a uniformly charged sphere is not kq/r^2. To determine the actual relationship, the easiest way is to use Gauss's Law. Are you familiar with Gauss's law?

3. Oct 1, 2008

### _F_

Yes I am familiar with Gauss's Law.

flux = integral(EAcos(theta)) = q_enclosed/e_0.

I'm still not sure where to go from here, though.

4. Oct 1, 2008

### gabbagabbahey

What is $Q_{enclosed}$ at r= 15cm?

5. Oct 1, 2008

### _F_

I know its not 14uC. It must be something else, but I can't seem to figure out what.

Q = (e_0)*E*A ?

A = 4(pi)r^2 = 4(pi)(.15m)^2 = .2827m^2 ?

6. Oct 1, 2008

### gabbagabbahey

Well the 14uC is distributed uniformly throughout the sphere, so what is the charge density of the sphere then?

7. Oct 1, 2008

### _F_

I would say Q/A = 14*10^-6 C * .2827 m^2 = 5*10^-5 C/m^2

But plugging that into: E = (5*10^-5 C/m^2)/e_0 = 5.59*10^6 N/C.

But my book says E = 1.21 MN/C

Last edited: Oct 2, 2008
8. Oct 2, 2008

### _F_

Can anyone point me in the right direction here?

I still can't wrap my head around this. :(

If the charge density is p = q/volume. Then p = (14*10^-6 C)/(4/3*pi*(.25m)^2 = 2.14*10^-4 C/m^3

But how do I use this?

9. Oct 2, 2008

### gabbagabbahey

Well, $Q_{enclosed}$ is the charge enclosed by your Gaussian surface. For part (a), the Gaussian surface does not enclose the entire volume; and so it does not enclose the entire charge. What volume is enclosed for r=15cm? Shouldn't the charge enclosed be that volume times the density that you just calculated?