Two point charges `q` and `-q` are separated by the distance `2l`. Find the flux of the electric field strength vector across a circle of radius `R`.

Explanation

Let us consider an elemental surface area as shown in the figure. Then flux of the vector `vec(E)` through the elemental area,
`d varphi = vec(E). Vec(dS) = E.dS = 2 E_(0) cos varphi dS (as vec(E) uarr uarr vec(dS))`
`= (2q)/(4pi epsilon_(0)(l^(2) + r^(2))) (l)/((l^(2) + r^(2))^(1//2)) (r d theta) dr = (2ql r dr d theta)/(4pi epsilon_(0) (r^(2) + l^(2))^(3//2))`
where `E_(0) = (q)/(4pi epsilon_(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 = (2ql)/(4pi epsilon_(0)) int_(0)^(R ) (r d r)/((r^(2) + l^(2))^(3//2)) int_(0)^(2pi) d theta`
`= (2q l xx 2pi)/(4pi epsilon_(0)) int_(0)^(R ) (r dr)/((r^(2) + l^(2))^(3//2)) = (q)/(epsilon_(0)) [1 - (1)/(sqrt(l^(2) + R^(2)))]`
It can also be solved by considering a ring elemeat or by using solid angle.