Question

The electric flux passing through the disc as shown in figure is approximately

• A. $0.5q/{ϵ}_0$
• B. $0.45q/{ϵ}_0$
• C. $0.30q/{ϵ}_0$
• D. $0.15q/{ϵ}_0$

Answer 1

Answer: (D)

Solid angle $\Omega$ = $2\pi (1-cos\theta )$
$\therefore \theta =45^o$
$\Omega =2\pi [1-\frac{1}{\sqrt{2}}]$
$\Omega =\sqrt{2}\pi (\sqrt{2}-1)$
$\frac{q}{{ϵ}_o}\to 4\pi$
$\varphi =\frac{q}{4\pi {\epsilon }_o}\sqrt{2}\pi \left(\sqrt{2}-1\right)$
$\varphi =\frac{\sqrt{2}\left(\sqrt{2}-1\right)q}{4{\epsilon }_o}=0.15q/{ϵ}_o$
= $\frac{0.15}{{ϵ}_o}q$