# Shouldn't the electric field in a solid insulating sphere be linear with radius?

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?

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!

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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

Hints:

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 $r$?

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Okay thank you very much I know see where I was wrong! –  Tom Granderson Oct 8 '12 at 21:24