The following picture represents two non-conductive spherical shells.

The charge $Q$ is distributed along its volume with the following density: $\rho(r) = C/r$ in region II. Calculate the variation of the magnitude of the charge $q(r)$ and the electric field $E(r)$ with $r$ being the distance from the center of the shells in the following regions:

- 1: $0<r< a$
- 2: $a\leq r \leq b$
- 3: $r > b$

After that, find the constant $C$.

Can someone please check if my approach and answers are correct? I do not have the correct answers for this one, and I do think that it's a very good problem for a beginner in electromagnetism.

Region 1:

If we create any gaussian sphere with radius $0<r<a$ very close to $a$, but less than a, we'll see that we do not have any charge inside this gaussian object, and hence, any kind of flux and because of that, any electric field. So for region 1 we'll have:

$E(r) = 0$ and $q(r) = 0$

Region 2:

$$Q = \frac{C}{r} V$$ $$ dQ = \frac{C}{r} dV$$ $$Q = \int dQ = \int \frac{C}{r} dV$$

But, summing surface areas times infinitely small $dr$ to get the volume of a sphere, hence:

$$dV = 4\pi r^2 dr$$

Substituting back:

$$Q= \int dQ = \int \frac{C}{r} 4\pi r^2 dr$$ $$Q= 4C\pi \int r$$

Since we are looking for charges inside any region 2, we get:

$$q(R)= 4C \pi \int_a^R r$$ $$q(R)= 2C \pi (R^2 - a^2)$$

Now, for the electric field, we can use gauss Law in a spherical gauss surface:

$$\oint E \cdot dA = \frac{q_{inside}}{\epsilon_0}$$ $$E 4\pi R^2= \frac{q(R)}{\epsilon_0}$$ $$E = \frac{2C\pi (R^2 - a^2)}{4\pi R^2 \epsilon_0}$$ $$E(R)= \frac{C (1 - \frac{a^2}{R^2})}{2 \epsilon_0}$$

Region 3:

Because $r$ is the distance from the center of the shells, and we are looking for $r>b$ we've got all the charge $Q$. So $q(r) = Q$ when $r>b$.

Now using the same reasoning as before, gauss Law:

$$\oint E \cdot dA= \frac{Q}{\epsilon_0}$$ $$E 4\pi R^2= \frac{Q}{\epsilon_0}$$ $$E(R)= \frac{Q}{4\pi R^2\epsilon_0}$$

Finding constant C:

$$Q= \int_{a}^{b} dQ = \int_a^b \frac{C}{r} 4\pi r^2 dr$$ $$Q= 4C\pi \int_{a}^{b} r dr$$ $$Q= 2C\pi (a^2-b^2)$$ $$C= \frac{Q}{2\pi(a^2-b^2)}$$

I would really appreciate any comment about my reasoning, and most important, to check if the answers are correct! Thanks a lot!

http://physics.stackexchange.com/questions/323501/spherical-shell-electric-field-and-charges