A hollow cylinder has charge ' $q$ ' $C$ within it. If ' $\phi$ ' is the electric flux associated with the…

A hollow cylinder has charge ' $q$ ' $C$ within it. If ' $\phi$ ' is the electric flux associated with the curved surface B , the flux linked with the plane surface A will be

1. $\frac{1}{2}\left(\frac{q}{\varepsilon_0}-\phi\right)$
2. $\frac{q}{2 \varepsilon_0}$
3. $\frac{\phi}{3}$
4. $\frac{q}{\varepsilon_0}-\phi$

Solution

As per Gauss' law the net flux through closed surface is, $\phi_{\mathrm{A}}+\phi_{\mathrm{B}}+\phi_{\mathrm{C}}=\frac{\mathrm{q}}{\varepsilon_0}$
Due to symmetry, the same flux passes through plane surfaces A and C, i.e., $\phi_{\mathrm{A}}=\phi_{\mathrm{C}}$ $\begin{array}{ll} \therefore \quad & 2 \phi_A+\phi=\frac{q}{\varepsilon_0} \\ & \phi_A=\frac{1}{2}\left(\frac{q}{\varepsilon_0}-\phi\right) \end{array}$ $\ldots\left(\text { Given: } \phi_B=\phi\right)$ /

Asked in: MHT CET 2024 (16 May Shift 2)

Practice more Electrostatics questions on Aicharya