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## Homework Statement

A volumetric charge density ##\rho = \frac{A}{r}## is distributed in the spheric region ##r_1 < r < r_2## (##A## is constant). A point charge ##q## is located in the center of the sphere (##r = 0##). What should be the value of ##A## so that the module of the electric field is constant for ##r_1 < r < r_2##?

## Homework Equations

Gauss Theorem for continuous distribution

##\Phi_s (\vec E_0) = \int_S \vec E_0 \cdot d\vec S = \frac{1}{\epsilon_0} \int_\tau \rho (x, y, z) d\tau##

## The Attempt at a Solution

What the problem is asking is to find ##A## so that ##E_0 = const##.

I started by finding the electric field for ##r_1## and ##r_2##, so, if I'm not wrong:

for ##r_1##

##E_0 4 \pi r_1^2 = \frac{1}{\epsilon_0} \int_\tau \rho (x, y, z) d\tau##

We know that ##dq = \rho (x, y, z) d\tau## so we have

##E_0 4 \pi r_1^2 = \frac{Q}{\epsilon_0}##

##E_0 = \frac{Q}{4 \pi \epsilon_0 r_1^2}## or ##E_0 = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r_1^2}##

same thing for ##r_2##

##E_0 = \frac{Q}{4 \pi \epsilon_0 r_2^2}## or ##E_0 = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r_2^2}##

Since we know that it's a sphere, ##Q = \rho \frac{4}{3} \pi r_1^3## and ##Q = \rho \frac{4}{3} \pi r_2^3##.

So the above expressions become

##E_0 = \frac{1}{4 \pi \epsilon_0} \frac{\rho}{r_1^2} \frac{4}{3} \pi r_1^3 = \frac{\rho r_1}{3 \epsilon_0}##

and

##E_0 = \frac{\rho r_2}{3 \epsilon_0}##

At this point I thought that by simply doing this ##\frac{\rho r_1}{3 \epsilon_0} = \frac{\rho r_2}{3 \epsilon_0}## I would have found ##A## but it's not true because I obviously find myself with ##r_1 = r_2##. How should I proceed doing this? Was it good finding the two electric field? Or I went the wrong way?