Question

A hollow cylinder has a charge q coulomb distributed uniformly within it. If $\varphi$ 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. $\frac{q}{{ϵ}_0}-\varphi$
• B. $\frac{1}{2}(\frac{q}{{ϵ}_0}-\varphi )$
• C. $\frac{q}{2{ϵ}_0}$
• D. $\frac{\varphi }{3}$

Answer 1

The correct option is B $\frac{1}{2}(\frac{q}{{ϵ}_0}-\varphi )$

$\text{Step 1: Total flux through cylinder using Gauss Law}$

By Gauss Law, total flux through the cylinder is given by:

${\varphi }_{total}=\frac{q}{{\epsilon }_0}$ $....\left(1\right)$

$\text{Step 2: Flux through plane surface A}$

As surfaces $A$ & $C$ is symmetrical so ${\varphi }_A={\varphi }_C$

${\varphi }_{total}=$ ${\varphi }_A+{\varphi }_B+{\varphi }_C=\frac{q}{{\epsilon }_0}$ [Using Equation $\left(1\right)$]

Given Flux through curved surface ${\varphi }_B=\varphi$

So,

$2{\varphi }_A+\varphi =\frac{q}{{\epsilon }_0}$

$\Rightarrow {\varphi }_A=\frac{1}{2}(\frac{q}{{\epsilon }_0}-\varphi )$