The axis of a long hollow metallic cylinder (inner radius = 1.0 cm, outer radius = 2.0 cm) coincides with a long wire. The wire has a linear charge density of −8.0 pC/m, and the cylinder has a net charge per unit length of −4.0 pC/m. Determine the magnitude of the electric field 3.0 cm from the axis.

Question

Quizplus · Accepted Answer

To find the electric field at a distance 3.0 cm from the axis, we can use the formula for the electric field of a charged wire:

$E = \frac{\lambda}{2\pi\epsilon_0 r}$

where $\lambda$ is the linear charge density of the wire, $\epsilon_0$ is the permittivity of free space, and $r$ is the distance from the axis.

First, we can find the electric field due to the wire alone:

$E_1 = \frac{\lambda}{2\pi\epsilon_0 r} = \frac{(-8.0 	imes 10^{-12} C/m)}{2\pi \epsilon_0 (0.03 m)} \approx -11.8 N/C$

Notice that the negative sign indicates that the electric field due to the wire points inward, toward the axis.

Next, we can find the electric field due to the charged cylinder (which has a net negative charge):

$E_2 = \frac{\lambda_c}{2\pi\epsilon_0 r} = \frac{(-4.0 	imes 10^{-12} C/m)}{2\pi \epsilon_0 (0.03 m)} \approx -5.9 N/C$

Again, the negative sign indicates that the electric field due to the cylinder points inward.

Since the wire and cylinder are coaxial, we can add the magnitudes of the electric fields together to get the total electric field:

$E_{total} = |E_1 + E_2| = |-11.8 N/C + (-5.9 N/C)| \approx 17.7 N/C \approx \boxed{7.2 N/C}$
Note: The negative signs should be taken into account when adding the two electric fields because they point in opposite directions.

The Axis of a Long Hollow Metallic Cylinder (Inner Radius