Question

Charge is distributed throughout a large cylinder metal of radius R. The charge per unit length in the cylindrical volume is λ. Use Gauss's law to find the magnitude of the electric field near the middle of the cylinder a distance d from the axis of the pipe, where d < R?

Quiz3-2015-Pic4
(2.pi.epsilon_0.lambda.R)
(2.pi.epsilon_0.lambda.d)
(lambda)/(2.pi.epsilon_0.d)
(lambda)/(2.pi.epsilon_0.R)
0

Answer 1

To find the electric field inside a charged cylindrical shell using Gauss's law, we can consider a Gaussian surface that is a cylinder of radius \( d \) (where \( d < R \)) and length \( L \), coaxial with the charged cylinder.

1. **Gauss's Law** states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space:

\[
\Phi_E = \frac{Q_{\text{enc}}}{\varepsilon_0}
\]

2. The total electric flux \( \Phi_E \) through the Gaussian surface is given by:

\[
\Phi_E = E(2\pi d L)
\]

where \( E \) is the magnitude of the electric field at the radius \( d \) from the axis.

3. Next, we need to determine the charge enclosed \( Q_{\text{enc}} \) within the Gaussian cylinder. The charge per unit length of the cylinder is \( \lambda \), and the length of our Gaussian cylinder is \( L \). Thus, the total charge enclosed by our Gaussian surface is:

\[
Q_{\text{enc}} = \lambda L
\]

4. Plugging this into Gauss's law, we have:

\[
E(2\pi d L) = \frac{\lambda L}{\varepsilon_0}
\]

5. The length \( L \) cancels from both sides:

\[
E(2\pi d) = \frac{\lambda}{\varepsilon_0}
\]

6. Solving for the electric field \( E \) gives:

\[
E = \frac{\lambda}{2\pi \varepsilon_0 d}
\]

Therefore, the magnitude of the electric field at a distance \( d \) from the axis of the cylinder (where \( d < R \)) is:

\[
E = \frac{\lambda}{2\pi \varepsilon_0 d}
\]

Thus, the correct answer is:

\[
\frac{\lambda}{2\pi \varepsilon_0 d}
\]