Question

A charge $(5\sqrt{2}+2\sqrt{5})$ coulomb is placed on the axis of an infinite disc at a distance a from the centre of disc. The flux of this charge on the part of the disc having inner and outer radius of a and 2a will be :

• A. $\frac{3}{2{\epsilon }_0}$
• B. $\frac{1}{2{\epsilon }_0}$
• C. $\frac{2[\sqrt{5}+\sqrt{2}]}{{\epsilon }_0}$
• D. $\frac{2\sqrt{5}+5\sqrt{2}}{2{\epsilon }_0}$

Answer 1

The correct option is A $\frac{3}{2{\epsilon }_0}$

As solid angle, $w=2\pi (1-cos\theta )$ where $\theta$ is half planer angle.

From trignometry, $tan{\theta }_1=\frac{a}{a}=1⟹{\theta }_1={45}^o$

Also $tan{\theta }_2=\frac{2a}{a}=2⟹cos{\theta }_2=\frac{1}{\sqrt{5}}$

Now solid angle subtended by shaded region, $w=w_2-w_1$

$w=2\pi (1-cos{\theta }_2)-2\pi (1-cos{\theta }_1)$ $=2\pi (1-\frac{1}{\sqrt{5}})-2\pi (1-\frac{1}{\sqrt{2}})$

$w=2\pi [\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{5}}]$

As flux through $4\pi$ steradian is equal to $\frac{q}{{ϵ}_o}$

Thus flux through $w$ steradian, $\varphi =\frac{q}{{ϵ}_o}\times \frac{w}{4\pi }$

$\varphi =\frac{2\sqrt{5}+5\sqrt{2}}{{ϵ}_o}\times \frac{2\pi [\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{5}}]}{4\pi }$

$\varphi =\frac{\sqrt{10}+5-2-\sqrt{10}}{2{ϵ}_o}$

$⟹\varphi =\frac{3}{2{ϵ}_o}$