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Introducing Flux

Two point cha...

Question

Two point charges q and -q are separated by a distance 2l, Find the flux of electric field strength vector across the circle of radius R placed with its centre coinciding with the midpoint of line joining the two charges in the perpendicular plane.

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Solution

$E = \frac{k q}{x^{2} + y^{2} + l^{2}} \frac{x^i + y^j - l^k}{\sqrt{x^{2} + y^{2} + z^{2}}} - \frac{k q}{x^{2} + y^{2} + l^{2}} \frac{x^i + y^j + l^z}{\sqrt{x^{2} + y^{2} + z^{2}}}$
$= - \frac{2}{4 π ϵ_{0}} \frac{q l^k}{(x^{2} + y^{2} + z^{2})^{3 / 2}}$
$ϕ = \int \to E . d \to A = \int E d x d y^k$
$= \frac{- 1}{2 π ϵ_{0}} q l \int \frac{d x d y}{(x^{2} + y^{2} + z^{2})^{3 / 2}}$
By doing change of coordinates,
$E = \frac{- 1}{2 π ϵ_{0}} q l \int \frac{r d θ d r}{(r^{2} + l^{2})^{3 / 2}}$
$= \frac{- 1}{2 π ϵ_{0}} q l 2 π \int_{0}^{R} \frac{r d r}{(r^{2} + l^{2})^{3 / 2}}$
By doing change of coordinates, $u = r^{2} + l^{2}$
$E = \frac{- q l}{ϵ_{0}} \int \frac{d u}{2 u^{3 / 2}}$
After solving this integral finally,
$E = \frac{q}{ϵ_{0}} (1 - \frac{1}{\sqrt{1 + (R / l)^{2}}})$

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