Question

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

• A. $0.5 \,q/ \epsilon _0$
• B. $0.45\, q/ \epsilon _0$
• C. $0.30\, q/ \epsilon _0$
• D. $0.15 \,q/ \epsilon _0$

Answer 1

Answer: (D)

Solid angle $\Omega$ = $2\pi(1-cos\theta)$
$\therefore \theta = 45^o$
$\Omega = 2\pi [1-\frac{1}{\sqrt{2}}]$
$\Omega = \sqrt2{\pi}(\sqrt{2}-1)$
$\frac{q}{\epsilon_o}\rightarrow 4\pi$
$\phi = \frac{q}{4 \pi \varepsilon_{o}} \sqrt{2}\pi\left(\sqrt{2}-1\right)$
$\phi =\frac{\sqrt{2}\left(\sqrt{2}-1\right)q}{4\varepsilon_{o}}=0.15 q/ \epsilon_o$
= $\frac {0.15}{\epsilon_o}q$