Solving Gauss's Law for Constant Electric Field: q, a, b

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To solve Gauss's Law for a constant electric field in the region defined by a < r < b, the charge density is given as ρ = A/r, where A is a constant. The problem involves a point charge q at the center of a spherical cavity, and the goal is to determine the value of A that ensures the electric field remains constant. The correct value of A is derived as A = q/(2πa²). A user attempted to apply Gauss's Law using a spherical Gaussian surface but encountered difficulties with the volume element calculation. Clarification was provided that the volume element should not be 2πr dr, indicating a need for careful consideration of the geometry in the calculations.
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Homework Statement


The spherical region a < r < b carries a charge per unit volume of \rho = A/r, where A is a constant. At the center (r = 0) of the enclosed cavity is a point charge q. What should be the value of A so that the electric field in the region a< r < b has constant magnitude?

The answer says A = q/(2pi a2)

2. The attempt at a solution
I drew a sphere with radius r as a Gaussian surface.
My work is in the attachment.

I got stuck from that point. I need help :!)
 

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You go wrong in the second line, the right-hand side: the volume element is NOT 2\pi r dr, but...

Otherwise, you got the right idea.
 
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