Total charge on an insulating sphere

[ehilge]

Homework Statement

A sphere of radius R=0.0850 m is made up of insulating material and has a spherically symmetric charge distribution. The radial component of the electric field inside the sphere is given by:

Er= (-5.20E4) r3, for r ≤ R,

where Er is in N/C when r is in meters and r is the distance from the center of the sphere.

What is the total charge Q on the entire sphere? (Hint: Gauss's Law.)

Homework Equations

Gauss's law:
$$\oint E\bullet dA$$ =Q_encl / epsilon₀

The Attempt at a Solution

I tried to set up a triple integral in spherical coords integrating -5.2E4 $$\rho$$³$$\rho$$ sin$$\phi$$ d$$\rho$$ d$$\phi$$d$$\theta$$

(btw, that's supposed to be rho cubed times rho)

from 0<=$$\rho$$<=.085, 0<=$$\theta$$<=2pi, and 0<=$$\phi$$<=pi

When I calculate this and multiply by epsilon, I get: -2.56E-12
which is a power of 10 off from the given answer which is: -2.56E-11

For some reason though, I feel like this is a coincidence and I'm missing something more than a power of ten somewhere. So, does my work seem to be right? If so where am I missing a power of 10 at? And also, I'm thinking there is an easier way to do this problem other than integrating so any insight into that would be great!
Thanks!

 

[ltjlogic]

I think it's interesting that you chose to use a triple integral. Here's what I would do:

To get the q(inside) all you have to do is integrate E * 4 pi r^2 * epsilon naught * dr. Remember to get the area in terms of dr now and not dA. Try it and see what you get!

**Edit** I've been trying to solve it now with your numbers, and am not getting the value that you listed above (-2.56E-11). At one point I get (-2.56 *10^-17), but that isn't close. Are all the numbers you provided correct?

 

[Doc Al]

To get the q(inside) all you have to do is integrate E * 4 pi r^2 * epsilon naught * dr. Remember to get the area in terms of dr now and not dA.

The flux is just E*Area = E(at r = R)*4 pi R^2.

 

[ehilge]

The flux is just E*Area = E(at r = R)*4 pi R^2.

That did it and is definitely easier than the way that I was trying to compute the answer, thanks!

And ltjlogic, your idea was what I had actually tried first. I'm not entirely sure why it doesn't work but I'm assuming it has something to with the fact the its measuring the charge on a line from the center of the sphere to r. Phi and theta are needed to complete the entire sphere. Not entirely sure on this reasoning though. Thanks for your suggestion.