Two point charges $q_{1}$ and $q_{2}$ are placed at a distance d apart as shown in the figure. The electric field intensity is zero at the point P on the line joining them as shown .
Write two conclusions that you draw from this.

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Solution

$Conclusion 1:$ Since, the net electric field at P is zero that means both electric field are equal and opposite. (One is rightwards and one is leftwards)

Therefore, both charges are of opposite sign.
$Conclusion 2:$
Electric field of a point charge is given by: $E = \frac{k q}{r^{2}}$
For equal electric field at point P i.e. $E_{1} = E_{2} \Rightarrow \frac{q_{1}}{r_{1}^{2}} = \frac{q_{2}}{r_{2}^{2}}$
As, $q_{1}$ is at more distance as compare to $q_{2}$ from point P i.e. $r_{1} \geq r_{2}$

Therefore, magnitude of $q_{1}$ is greater than $q_{2}$ i.e. $q_{1} \geq q_{2}$

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