Question

A point charge $q$ is placed at a point on the axis of a non-conducting circular plate of radius $r$ at a distance $R(R>>r)$ from its center. The electric flux associated with the plate is :

• A. $\frac{q{r}^2}{4\pi {\epsilon }_0{R}^2}$
• B. $\frac{q{r}^2}{4{\epsilon }_0{R}^2}$
• C. $\frac{q{r}^2}{4{\epsilon }_0{r}^2}$
• D. $\frac{q}{4{\epsilon }_0}$

Answer 1

The correct option is D $\frac{q}{4{\epsilon }_0}$

Charge $=q$

radius $=r$

$R>>r$

its centre. The electric flux associated with the plate $E_x$ on the ring of radius $X=\frac{KQ\cos \theta }{(b^2+x^2)}$

$E_x=\frac{KQb}{{(b^2+x^2)}^{3/2}}$

$\phi =\varphi \int E_{x}da$

$\phi ={\int }_{x=0}^{x=R}\frac{KQ{b}_2\pi xdx}{{(b^2+x^2)}^{3/2}}$

putting $x=b\tan \theta$

$\phi =\frac{2\pi KQb}{b^3}{\int }_0^{{\tan }^{-1}\left(\frac{R}{b}\right)}\frac{b\left(\tan \theta \right)b{\sec }^2\theta }{{\sec }^3\theta }d\theta$

$\phi =2\pi KQ{\int }_0^a\sin \theta d\theta$ $[\alpha ={\tan }^{-1}\sqrt{3}={60}^0]$

$\phi =2\pi KQ(1-\cos \alpha )$

$\phi =\frac{Q}{2{ϵ}_0}\times \frac{1}{2}=\frac{Q}{4{ϵ}_0}$