A long cylinder (radius = 3.0 cm) is filled with a nonconducting material which carries a uniform charge density of 1.3 μC/m³. Determine the electric flux through a spherical surface (radius = 2.0 cm) which has a point on the axis of the cylinder as its center.

Question

Quizplus · Accepted Answer

First, we can find the charge contained within a length of the cylinder:
$q=\lambda \cdot l$
$q=(1.3\cdot10^{-6} C/m^3) \cdot (\pi \cdot (3.0 cm)^2 \cdot (2.0 cm))$
$q=23.34 \cdot 10^{-6}C$
Next, we can use Gauss's Law to find the electric flux through the spherical surface:
$Φ_{E} = EA$ where $E$ is the electric field and $A$ is the area of the sphere.
We can use symmetry to claim that the electric field will be radial and constant on the surface of the sphere, so 
$E=\frac{q}{4\pi \epsilon_0 r^2}$.
$Φ_{E} = \frac{q}{4\pi \epsilon_0 r^2} \cdot A$
$Φ_{E} = \frac{q}{r^2 \epsilon_0}$
$Φ_{E} = \frac{23.34 \cdot 10^{-6}C}{(2.0 cm)^2 \cdot 8.85 \cdot 10^{-12} N^{-1}C^2m^{-2}}$
$Φ_{E} = 4.9N \cdot m^{-2}C$
$Φ_{E} = 4.9N \cdot m^{-1}C$ (since area of a sphere is $4\pi r^2$)
$Φ_{E} = 4.9N \cdot m \cdot C^{-1}$
Therefore, the answer is B.

A Long Cylinder (Radius = 3