# Electric field calculation

## Homework Statement

We have a sphere with radius b, within that sphere there's another sphere with radius a. Between the two spheres we have an electric charge with density A/r. Also, we have a charge Q in the center. We need to find the constant A so that the field between a and b is independent of r (meaning, it's constant for a<=r<=b).

Gauss' law.

## The Attempt at a Solution

I added the scanned pages I've written, it should be understandable. As I said, I used Gauss' law to calculate the field, using a sphere with radius r (a<=r<=b) to calculate the electric flux. After finding the electric field as a function of r, I calculated d(E(r))/dr and demanded that it would be=0, but I can't seem to get a solution for A that's independent of r.

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collinsmark
Homework Helper
Gold Member

$$Qin = \frac{A}{r} \left[ \frac{4}{3}\pi r^3 - \frac{4}{3}\pi a^3 \right] + Q$$

That's not right. You're going to need to re-evaluate the volume integral. The $$(4/3)\pi r^3$$ terms only makes sense if the charge distribution is uniformly distributed about the sphere. You can't just multiply them by A/r. It doesn't work that way. You'll have to re-do the volume integral, integrating A/r' from r' = a to r in spherical coordinates (be careful to use the correct volume differential when performing the integral -- it's not simply dr') .

Once you have an expression for the electric field, E, don't worry about taking the derivative. Instead, find the electric field at r = a, and call that Ea. Then find Eb, the electric field at r = b. Set Ea and Eb equal to each other and solve for A.

$$Qin = \frac{A}{r} \left[ \frac{4}{3}\pi r^3 - \frac{4}{3}\pi a^3 \right] + Q$$
That's not right. You're going to need to re-evaluate the volume integral. The $$(4/3)\pi r^3$$ terms only makes sense if the charge distribution is uniformly distributed about the sphere. You can't just multiply them by A/r. It doesn't work that way. You'll have to re-do the volume integral, integrating A/r' from r' = a to r in spherical coordinates (be careful to use the correct volume differential when performing the integral -- it's not simply dr') .