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?

Answer 1

To find the electric field inside a uniformly charged cylinder using Gauss's law, we will take advantage of the symmetry of the problem. Given that the cylinder has a uniform linear charge density \(\lambda\) (charge per unit length), we will consider a coaxial Gaussian surface in the form of a cylindrical shell of radius \(d\) and length \(L\), where \(d < R\).

Step 1: Write down Gauss's Law

Gauss's law states that:

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

where \(\Phi_E\) is the electric flux through our Gaussian surface, \(Q_{\text{enc}}\) is the charge enclosed by the Gaussian surface, and \(\varepsilon_0\) is the permittivity of free space.

Step 2: Calculate the charge enclosed \(Q_{\text{enc}}\)

Since we have a cylindrical charge distribution with charge density \(\lambda\) per unit length, the charge enclosed by the Gaussian surface is given by:

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

where \(L\) is the length of the Gaussian cylinder.

Step 3: Find the electric flux \(\Phi_E\)

The electric field \(E\) inside the cylinder will be radially symmetric around the axis; thus, the electric field is constant at distance \(d\) from the axis of the cylinder. The total flux through the Gaussian surface is then:

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

where \(2\pi d L\) is the lateral surface area of the cylindrical shell.

Step 4: Apply Gauss's Law

Substituting the expressions for \(\Phi_E\) and \(Q_{\text{enc}}\) into Gauss's law gives:

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

Step 5: Solve for the electric field \(E\)

We can simplify the equation by canceling \(L\) from both sides (assuming \(L \neq 0\)):

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

Now, solving for \(E\):

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

Final Result

Thus, the magnitude of the electric field a distance \(d\) from the axis of a uniformly charged cylinder (where \(d < R\)) is given by:

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

This result shows that the electric field inside a uniformly charged infinite cylinder varies inversely with the distance \(d\) from the axis.