Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am a senior in High School who is taking the course AP Physics Electricity and Magnetism.

I was studying Gauss's laws and I found this problem:

A solid insulating sphere of radius R contains a positive charge that is distrubuted with a volume charge density that does not depend on angle but does increase with distance from the sphere center. Which of the graphs below correctly gives the magnitude E of the electric field as a function of the distance r from the center of the sphere?

Choices A through E

The correct answer is given to be choice D but I cannot see why the answer is D. Isn't the equation for electric field in this case just $E = \frac{q \cdot r}{4 \pi \epsilon _{0} \cdot R^{3} } $ if $r \le R$?

This formula occurs for spherical insulators as given by the textbook Fundamentals of Physics by Halliday/Resnick. According to this equation for the electric field, the graph should clearly be linear until $r=R$.

That is why I think the answer is C. I believe this is a problem with the textbook, am I correct?

If I am wrong can someone please explain why I am wrong?

Thank you very much for your time!

share|cite|improve this question
You would be correct if the charge density was uniform, but the problem states that it increases with radius. – user2963 Oct 8 '12 at 21:07
@zephyr that sounds like answer material – David Z Oct 8 '12 at 21:07
Please forgive me I am fairly new to Gauss's law but if the object is uniform then isnt the correct answer B because it is unchanging? – Tom Granderson Oct 8 '12 at 21:13
up vote 2 down vote accepted


1) The key sentence is

the volume charge density $\rho$ [...] does increase with distance from the sphere center.

2) From Gauss' law in integral form $\Phi_E=\frac{Q}{\epsilon}$, one gets

$$\tag{1} 4\pi r^2 \cdot E(r)~=~ \frac{1}{\epsilon}\int_0^r\! 4\pi r^{\prime 2} dr^{\prime} ~ \rho(r^{\prime}). $$

3) To get the idea, say for simplicity that the increase is linear

$$\tag{2} \rho(r)~\propto~r\qquad \text{for}\qquad r~\leq~ R. $$

4) Use eqs. (1) and (2) to prove that then the electric field increases quadratically

$$\tag{3} E(r)~\propto~r^2\qquad \text{for}\qquad r~\leq~ R. $$

5) What happens if $\rho(r)=Ar^{\alpha}$ is a power law of $r$?

share|cite|improve this answer
Okay thank you very much I know see where I was wrong! – Tom Granderson Oct 8 '12 at 21:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.