Homework Help: Gauss/Electric Field question

1. Feb 14, 2004

paul11273

Here is a HW question that I have been wrestling with. Hopefully I have gone about solving it correctly. There is no answer in the back of the text to check my result against. Can anyone have a look and confirm?
Question-
"A solid plastic sphere with radius of 10.0cm has a charge with uniform density throughout it's volume. The electric field 5cm from the center is E=86.0 kN/C radially inward. Find the magnitude of the electric field 15cm from the center."

First I figured I needed to find the charge density. I set EA=qr/e0 ,where e0 is permittivity of free space and qr is total charge at radius of .05m given in the problem. I get to E=pr/3e0 where I am using p as charge density.
I solved for p with the given E and r, so p=4.6*10^-5 (C/m^3).

Next, I found the total charge within the .05m sphere with Q=pV. I got 1.91*10^-7 (C).

Lastly, I used E=ke(Q/r^2) to find the electrical field at radius=.15m. My final result is E=7.63*10^4 (N/C).

Can anyone tell me if I am correct? If not, where did I go wrong? Any help would be appreciated. I feel like I may have done too much work, there is probably a shorter way to the answer.
Thanks.

2. Feb 15, 2004

Staff: Mentor

You need to find the total charge, which extends to radius = 0.1m.
If you had the correct "Q", that method would work.
Here's how I would do this problem. First, I would express all distances in terms of the first radius given, R = 0.05. Thus the radius of the sphere of charge would be 2R; the distance at which you are to find the field would be 3R.

Then I'd apply Gauss's law find the charge density (like you did), but I would express it in terms of the field given (call it E1 and R.

Then I'd find the total charge (like you intended to do) for the sphere of radius 2R. Don't plug in numbers, just get an expression.

Then I'd use Gauss's law again to find the field at a distance of 3R, using the Q I just found.

Do it this way and you'll find that things cancel and you get a simple expression of the field at 3R in terms of the original field at R. The moral: don't be in a rush to plug in numbers and do arithmetic.

3. Feb 15, 2004

paul11273

Thanks, Doc Al, for taking the time to review my work.

After checking my work I realize that I actually did find the charge of the .1m radius, not the .05m. I must have made an error typing it in. Good catch though.

I will try reworking this problem with your method of using R, 2R and 3R. I understand the concept, but I won't really get it until I work it all out on paper.

Thanks again.