The field just outside the surface of a long conducting cylinder which has a 2.0-cm radius points radially outward and has a magnitude of 200 N/C. What is the charge density on the surface of the cylinder?

Question

Quizplus · Accepted Answer

The electric field just outside the surface of a cylinder with radius $r$ and linear charge density $\lambda$ is given by $E=\frac{\lambda}{2\pi\epsilon_0 r}$.
In this case, $r=2$ cm and $E=200$ N/C. Using $2\pi\epsilon_0=8.85	imes 10^{-12}$ Nm$^2$/C$^2$, we can solve for $\lambda$:
$$\lambda=2\pi\epsilon_0 r E = 2\pi(8.85	imes 10^{-12}	ext{ Nm}^2/	ext{C}^2)(0.02	ext{ m})(200	ext{ N/C}) \approx 1.77	imes 10^{-9}	ext{ C/m}$$
To convert to nC/m, we multiply by $10^9$, giving a final answer of $\boxed{	extbf{(B) }1.8}$ nC/m.

The Field Just Outside the Surface of a Long Conducting