# Electric charge in a shell

## Homework Statement

(Note: Don't worry about significant digits. I just want to be able to do the question and will worry about significant digits on the exam.)

In the figure below, a non-conducting spherical shell of inner radius a=2.00cm and outer radius b=2.40cm has (within its thickness) a positive volume density $$\rho$$=A/r, where A is a constant and r is the distance from the centre of the shell. In addition, a small ball of charge q=45.0 fC is located at the centre.

What value should A have if the electric field in the shell (a$$\leq$$r$$\leq$$b) is to be uniform?

(See attached file for the picture. sorry it isn't very good. That's a "q" beside the +ve charge in the middle.)

## Homework Equations

Charge density : $$\rho= \frac{dq}{dV}$$

Volume : $$V = 4\pi r^3$$

Electric field : $$E = \frac{1}{4\pi \epsilon} \int \frac{dq}{r^2}$$

## The Attempt at a Solution

E at a distance r anywhere within [a, b] is constant.
E is caused by the point charge and distribution of charges in the volume.

$$E_{net}$$ is constant, therefore

$$E_{charge}=-E_{volume}$$

$$\frac{1}{4\pi \epsilon} \int \frac{dq_{charge}}{r_{charge}^2} = - \frac{1}{4\pi \epsilon} \int \frac{dq_{volume}}{r_{volume}^2}$$

the E for a point charge doesn't need to have an integral, because it works out that way.

$$\frac{q_{charge}}{r^2} = - \int \frac{dq_{volume}}{r_{volume}^2}$$

Charge density : $$\rho= \frac{A}{r} = \frac{dq}{dV}$$

$$dq = \frac{A}{r} dV$$

$$\frac{q_{charge}}{r^2} = - \int \frac{A dV}{r^3}$$

dV is related to r because $$V = \frac{4}{3} \pi (r^3 - a^3)$$

then $$r^3 = \frac{3V + 4 \pi a^3}{4 \pi}$$

$$\frac{q_{charge}}{r^2} = - \int \frac{4 \pi A dV}{3V + 4 \pi a^3}$$

$$\frac{q_{charge}}{r^2} = - 4 \pi A \int \frac{dV}{3V + 4 \pi a^3}$$

$$\frac{q_{charge}}{r^2} = - \frac{4 \pi A}{3} ln (3V + 4 \pi a^3)$$

I don't know what to do after this, and I don't know what to do. I don't know if I messed up something.

This may be supposed to be done with Gauss' Law, but I couldn't figure out what to do with that.

Gauss' Law : $$\Phi = \oint E da$$ (In this case since E and a are in the same direction, the dot product just became E da)

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