# Gauss' Law: Solid Non-conducting Cylinder

by Bryon
Tags: cylinder, gauss, nonconducting, solid
 P: 123 Hi, Bryon. First of all, we have Gauss' law: $$\oint_\mathcal{S} \mathbf{E} \cdot d\mathbf{S} = \frac{Q_\text{int}}{\epsilon_0}$$ The point where they ask you to get the E field is inside the cylinder, so to get this E field, you would need to set up a gaussian cylinder of radius 7 cm (I'd work symbolically until the very end) and use Gauss' law. The evaluation of the LHS is very easy, it is merely the E times the curved outer surface area of the gaussian cylinder (the sides don't add to the flux integral because there, E is perpendicular to the surface). Assuming that the cylinder is of radius $$r$$ and length $$L$$, this turns out to be $$E 2 \pi r L$$. The RHS is slightly trickier, as this involves the total charge contained in the gaussian cylinder. That is: $$Q_\text{int} = \int_V \rho dv$$ To work out the differential element of volume, we remember that the volume of a cylinder is $$v = \pi r^2 L$$, and differentiating with respect to $$r$$, we have $$dv = 2 \pi r L dr$$. We also have the charge density $$\rho$$ as a function of $$r$$. Plugging it into the charge equation, $$Q_\text{int} = \int_V \rho dv = \int_0^r A r'^2 2 \pi r' L dr' = \frac{A \pi L r^4}{2}$$ where I added $$'$$ to the integrand to not confuse it with the upper limit. All you need to do now is set both sides of Gauss' law equal to each other, solve for the E field, and not forgetting to do unit conversion, evaluate it at r = 7 cm. Notice how when you set them equal, the $$L$$'s cancel out, as expected - this is a better method than to assume it's value to be 1. 1 What? cm? m? Good luck.