Question

A point charge $Q$ is located on the axis of a disc of radius $R$ at a distance $b$ from the plane of the disc. If one-fourth of the total electric flux from the charge passes through the disc, then choose the correct option.

• A. $R=\sqrt{3}b$
• B. $\sqrt{3}R=b$
• C. $2R=\sqrt{3}b$
• D. $R=\frac{2b}{\sqrt{3}}$

Answer 1

The correct option is A $R=\sqrt{3}b$

From the concept of solid angle we know that, solid angle subtended by disc is $2\pi (1-\cos \theta )$

Since the flux through the solid angle $4\pi$ is
${\varphi }_{\text{total}}=\frac{Q}{{\epsilon }_0}...\left(1\right)$
So, the flux through disc with solid angle $2\pi (1-\cos \theta )$ will be

${\varphi }_{disc}=\frac{2\pi (1-\cos \theta )Q}{{\epsilon }_0\left(4\pi \right)}$

$\Rightarrow {\varphi }_{disc}=\frac{(1-\cos \theta )Q}{2{\epsilon }_0}...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we have,
${\varphi }_{\text{disc}}=\left({\varphi }_{\text{total}}\right)\left(\frac{1-\cos \theta }{2}\right)\dots \dots \left(3\right)$

According to question, ${\varphi }_{\text{disc}}=\frac{1}{4}\left({\varphi }_{\text{total}}\right)\dots \dots \left(4\right)$

From equation $\left(3\right)$ and $\left(4\right)$ we get,
$\frac{1}{4}=\left(\frac{1-\cos \theta }{2}\right)$
$\Rightarrow \cos \theta =\frac{1}{2}$
$\therefore \theta ={60}^{\circ }$

From above figure,
$\frac{R}{b}=\tan {60}^{\circ }=\sqrt{3}$
$\therefore R=\sqrt{3}b$

Hence option (a) is correct answer.