Question

A hollow cylinder has a charge $q\text{C}$ within it placed at the centre. If the electric flux associated with the curved surface $B$ is $\varphi$, the flux linked with the plane surface $A$ 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 $\frac{1}{2}(\frac{q}{{ϵ}_0}-\varphi )$

Given that,
The flux associated with the surface $B={\varphi }_B=\varphi$
The total flux (${\varphi }_{total}$) linked with the cylinder is the sum of the flux through the faces $A,B$ and $C$.
Let the flux through the face $A,B$ and $C$ is ${\varphi }_A,{\varphi }_B$ and ${\varphi }_C$
${\varphi }_{total}={\varphi }_A+{\varphi }_B+{\varphi }_C\dots \left(i\right)$

The face $A$ and $C$ is symmetric with respect to the charge $q$, so flux passing through them will be same.
$\Rightarrow {\varphi }_A={\varphi }_C$
According to Gauss law,
${\varphi }_{total}=\frac{q}{{ϵ}_0}$
$\Rightarrow {\varphi }_{total}=2{\varphi }_A+{\varphi }_B$
$\Rightarrow \frac{q}{{ϵ}_0}=2{\varphi }_A+{\varphi }_B$
$\Rightarrow 2{\varphi }_A=\frac{q}{{ϵ}_0}-{\varphi }_B$
$\Rightarrow {\varphi }_A=\frac{1}{2}(\frac{q}{{ϵ}_0}-\varphi )$

Hence, option (b) is correct.


$\begin{array}{c}\text{Why this question}?\\ \text{Tip: The surface located at symmetric}\\ \text{position w.r.t to charge. The charge}\\ \text{will have equal linked flux, because}\\ \text{equal number of field lines will be}\\ \text{passing through these surfaces}\end{array}$