Question

A charge $Q$ is placed symetrically at the centre of a closed cylinder as shown in the figure. The flux associated with the curved surface of the cylinder is

• A. $\frac{Q}{2{\epsilon }_0}\cos \alpha$
• B. $\frac{Q}{{\epsilon }_0}\cos \alpha$
• C. $\frac{2Q}{{\epsilon }_0}\cos \alpha$
• D. $\frac{Q}{{\epsilon }_0}\sin \alpha$

Answer 1

The correct option is B

$\frac{Q}{{\epsilon }_0}\cos \alpha$

$\text{Flux through curved surface = Total flux through cylinder-Flux through caps A and B}$

${\varphi }_c={\varphi }_{total}-({\varphi }_A+{\varphi }_B)...\left(1\right)$

Using formula for flux through cone of semi angle $\alpha$, flux through caps $\text{A}$ and $\text{B:}$

${\varphi }_A={\varphi }_B=\frac{Q}{2{\epsilon }_0}(1-\cos \alpha )$

Total flux through closed
cylinder, ${\varphi }_{total}=\frac{Q}{{\epsilon }_0}$

So, from $\left(1\right)$ flux through curved surface, ${\varphi }_c=\frac{Q}{{\epsilon }_0}-\frac{2Q}{2{\epsilon }_0}(1-\cos \alpha )$

$\Rightarrow {\varphi }_c=\frac{Q}{{\epsilon }_0}(1-1+\cos \alpha )$

$\therefore {\varphi }_c=\frac{Q}{{\epsilon }_0}\cos \alpha$

Hence, option (b) is correct answer.