Question

Two point charges $q$ and $-q$ are separated by a distance $2a$ (Fig.). Evaluate the flux of electric field strength vector across a circle of radius $R$.

Answer 1

Let us consider an elemental surface area as shown in the figure. Then flux of the vector $\overrightarrow{E}$ through the elemental area,

$d\Phi =\overrightarrow{E}\cdot d\overrightarrow{S}=EdS=2E_0\cos \phi dS(as\overrightarrow{E}\uparrow \uparrow d\overrightarrow{S})$

$=\frac{2q}{4\pi {ϵ}_0(l^2+r^2)}\frac{l}{(l^2+r^2{)}^{1/2}}\left(rd\theta \right)dt=\frac{2qlrdrd\theta }{4\pi {ϵ}_0(r^2+l^2{)}^{3/2}}$

where $E_0=\frac{q}{4\pi {ϵ}_0(l^2+r^2)}$ is magnitude of field strength due to any point charge at the point of location of considered elemental area.

Thus $\Phi =\frac{2ql}{4\pi {ϵ}_0}{\int }_0^R\frac{rdr}{(r^2+l^2{)}^{3/2}}{\int }_0^{2\pi }d\theta$

$=\frac{2ql\times 2\pi }{4\pi {ϵ}_0}{\int }_0^R\frac{rdr}{(r^2+l^2{)}^{3/2}}=\frac{q}{{ϵ}_0}[1-\frac{l}{\sqrt{l^2+R^2}}]$

It can also be solved b considering a ring element or by using solid angle.