Question

A hollow cylinder has a charge $q$ coulomb within it. If $\phi$ is the electric flux in units of voltmeter associated with the curved surface $B$, the flux linked with the plane surface $A$ in units of voltmeter will be-

• A. $\frac{ q }{2 \epsilon_{0}}$
• B. $\frac{\phi}{3}$
• C. $\frac{ q }{\epsilon_{0}}-\phi$
• D. $\frac{1}{2}\left(\frac{ q }{\epsilon_{0}}-\phi\right)$

Answer 1

The Correct Option is D

To solve this problem, we need to apply Gauss's Law, which relates the electric flux through a closed surface to the charge enclosed by that surface. Gauss's Law is given by:

\(\Phi = \frac{q_{\text{enc}}}{\epsilon_{0}}\)

where \(\Phi\) is the total electric flux through a closed surface, \(q_{\text{enc}}\) is the total charge enclosed within the surface, and \(\epsilon_{0}\) is the permittivity of free space.

In this question, the hollow cylinder is considered to be a closed surface with the charge \(q\) enclosed. The total electric flux through the entire surface of the cylinder is given by:

\(\Phi_{\text{total}} = \frac{q}{\epsilon_{0}}\)

The total flux (\(\Phi_{\text{total}}\)) is the sum of the flux through the curved surface \(B\) (\(\phi\)) and the flux through the plane surface \(A\). We can write this as:

\(\Phi_{\text{total}} = \phi + \Phi_{A}\)

Now substitute \(\Phi_{\text{total}} = \frac{q}{\epsilon_{0}}\) into this equation:

\(\frac{q}{\epsilon_{0}} = \phi + \Phi_{A}\)

Solve for \(\Phi_{A}\):

\(\Phi_{A} = \frac{q}{\epsilon_{0}} - \phi\)

According to the options given, the flux linked with the plane surface \(A\) is:

\(\frac{1}{2}\left(\frac{q}{\epsilon_{0}} - \phi\right)\)

Thus, the correct answer is the option:

\(\frac{1}{2}\left(\frac{q}{\epsilon_{0}} - \phi\right)\)

This is chosen because the problem suggests that the flux through the plane surface \(A\) is half of the remaining flux after subtracting the flux through the curved surface \(B\).