Question

A charge $Q$ is placed symetrically at the centre of a closed cylinder as shown in the figure. If $\alpha ={60}^{\circ }$ then the flux associated with the curved surface of the cylinder will be

• A. $\frac{Q}{{2\epsilon }_0}$
• B. $\frac{Q}{{\epsilon }_0}$
• C. $\frac{2Q}{{\epsilon }_0}$
• D. $\frac{Q}{{4\epsilon }_0}$

Answer 1

The correct option is A $\frac{Q}{{2\epsilon }_0}$

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

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

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

${\varphi }_A={\varphi }_B=\frac{\frac{Q}{{\epsilon }_0}}{4\pi }\times \text{Solid angle}$

${\varphi }_A={\varphi }_B=\frac{\frac{Q}{{\epsilon }_0}}{4\pi }\times 2\pi (1-\cos \alpha )$

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

Total flux through closed cylinder:

${\varphi }_{total}=\frac{Q}{{\epsilon }_0}$

So, from $\left(1\right)$ flux through curved surface:

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

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

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

Given, $\alpha ={60}^{\circ }$,

$\Rightarrow {\varphi }_c=\frac{Q}{{\epsilon }_0}\cos {60}^{\circ }=\frac{Q}{{2\epsilon }_0}$

Hence, option (a) is the correct answer.