A long cylindrical shell (radius = 2.0 cm) has a charge uniformly distributed on its surface. If the magnitude of the electric field at a point 8.0 cm radially outward from the axis of the shell is 85 N/C, how much charge is distributed on a 2.0-m length of the charged cylindrical surface?
A) 0.38 nC
B) 0.76 nC
C) 0.19 nC
D) 0.57 nC
E) 0.98 nC

Question

Quizplus · Accepted Answer

The electric field $E$ outside a uniformly charged cylindrical shell is given by $E = \frac{\lambda}{2\pi\epsilon_0 r}$, where $\lambda$ is the linear charge density ($\frac{Q}{L}$, with $Q$ being the total charge and $L$ the length of the cylinder), $r$ is the radial distance from the axis of the cylinder (8.0 cm or 0.08 m in this case), and $\epsilon_0$ is the vacuum permittivity ($8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2$). Rearranging for $Q$, we get $Q = E \cdot 2\pi\epsilon_0 r \cdot L$. Substituting the given values ($E = 85 \, \text{N/C}$, $r = 0.08 \, \text{m}$, $L = 2.0 \, \text{m}$), we find $Q = 85 \cdot 2\pi \cdot 8.85 \times 10^{-12} \cdot 0.08 \cdot 2.0$, which calculates to approximately $0.76 \times 10^{-9}$ C or 0.76 nC.

A Long Cylindrical Shell (Radius = 2