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Spherical shell - Electric field and charges

1 vote

The following picture represents two non-conductive spherical shells.

Two 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!

asked Apr 5, 2017 in Physics Problems by breisfm (110 points)
edited May 14, 2018 by sammy gerbil
Your text formatting has gone wrong and is consequently difficult to read. There are still some bugs in the site so it may not be your fault. I notice you posted on Physics SE and your question was closed. Your text was fine there, and I suppose you copied it.

Looking over your post on Physics SE, you seem to know what you are doing, and don't seem to have any doubts about it. Is there any reason why you think your answers might be wrong? Personally I do not like checking someone else's calculations unless there is a good reason. It is tedious, and unlikely to uncover a mistake.
Hi Sammy. Yes, I copied because it was closed there. I agree with you about being tedious to check someone else's calculations. The thing is that I do not have the answer for this question. I think that I'm correct, but I don't have 100% sure because I do not have the book with the answer. Anyway, I'm new to the website and I don't know in what way should I format the text. I was using LaTex notation in stackexchange... Thank you for commenting!

1 Answer

0 votes

If I understand correctly, the only charge is in region II. The element of volume is a spherical shell of width $dr$ and volume $dV=4\pi r^2 dr$. The charge inside a sphere of radius $r$ in region II is $$Q(r)=\int_a^r \rho dV=4\pi C \int_a^r \frac{1}{r}r^2dr=4\pi C (r-a)$$ The total charge in region II is $$Q=Q(b)=4\pi C(b-a)$$ So we can write $$Q(r)=\frac{(r-a)}{(b-a)}Q$$

Applying Gauss' Law the electric fields in the three regions are :

  • region I : $E=0$ since no charge is enclosed
  • region II : $E=\frac{Q(r)}{4\pi \epsilon_0 r^2}=\frac{(r-a)}{(b-a)}\frac{Q}{4\pi \epsilon_0 r^2}$
  • region III : $E=\frac{Q}{4\pi \epsilon_0 r^2}$

Check: the electric field is continuous at the boundaries.

answered May 14, 2018 by sammy gerbil (28,806 points)