Suppose we have a cylindrical shell with radius $R$ and length $L$ that has a uniform charge distribution with total charge $Q$. The cylinder does not have bases, so the charge is only distributed on the wall that wraps around the cylinder at the radius $R$. What is the electric field at a point $P$, which is a distance $z$ from the center of the cylinder, along the axis that passes through the center of the cylinder and parallel to its wall? What happens to the electric field as $z = 0$? What about for very large $z$? Why?

• The point $P$ is a distance $z$ away from the center of the cylinder.
• The cylinder has a charge $Q$, which is uniformly distributed.
• The cylinder has length $L$ and radius $R$.
• The electric field due to a ring with the same dimensions in the $xy$-plane at a point along the $z$-axis is $$\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{Qz}{(R^2+z^2)^{3/2}}\hat{z}$$ We found this out in the previous example.

• Find the electric field at $P$.

• This is a perfect cylinder: This simplifies down the geometry of the problem and allows us to use any equations related to the geometry of the cylinder
• There is no top or bottom: We make this assumption so that we can break the cylinder up into rings and not have to do anything with the top or bottom sides of the cylinder (essentially we are dealing with a 2D tube).