Avery long insulating cylinder has radius r and carries positive charge distributed throughout its volume. the charge distribution has cylindrical symmetry but varies with perpendicular distance from the axis of the cylinder. the volume charge density is rho(r)=α(1−r/r), where α is a constant with units c/m^3 and r is the perpendicular distance from the center line of the cylinder.
1. derive an expression, in terms of α and r, for e(r), the electric field as a function of r.
2. do this for rr.
3. do your results agree for r=r?

1.

2.

3.The results from part 1 and 2 agree when r = R.

Explanation:

The volume charge density is given as

We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.

1. Since the cylinder is very long, Gauss’ Law can be applied.

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

where ‘h’ is the length of the imaginary Gaussian surface.

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

3. At the boundary where r = R:

As can be seen from above, two E-field values are equal as predicted.

i think it’s a. this is because if they measured the mass of something with a mass of 150g, then their final answer would have to be approximately 150g. though all of their answers are almost 150, rosa and tran have the closest answers (or the exact answer) to the accurate one.

it has to be b, about 97 percent.