Two point charges +10 q and -4 q are located | vedclass

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Question

(A) Let the point where the net electric field is zero be at a distance $r$ to the of the charge $-4 q$ (at $x=L$).
At this point, the distance from

Vedclass · Accepted Answer

$r = \frac{\sqrt{2}}{\sqrt{5}-\sqrt{2}} L$ to point $B$

Vedclass · Answer

(A) Let the point where the net electric field is zero be at a distance $r$ to the of the charge $-4 q$ (at $x=L$).
At this point, the distance from $+10 q$ (at $x=0$) is $(L+r)$.
The electric field due to $+10 q$ is $E_1 = \frac{K(10 q)}{(L+r)^2}$ and due to $-4 q$ is $E_2 = \frac{K(4 q)}{r^2}$.
For the net electric field to be zero, $E_1 = E_2$.
$\frac{10 q}{(L+r)^2} = \frac{4 q}{r^2}$
$\frac{\sqrt{10}}{L+r} = \frac{2}{r}$
$\frac{\sqrt{5} \cdot \sqrt{2}}{L+r} = \frac{\sqrt{2} \cdot \sqrt{2}}{r}$
$\frac{\sqrt{5}}{L+r} = \frac{\sqrt{2}}{r}$
$\sqrt{5} r = \sqrt{2} L + \sqrt{2} r$
$r(\sqrt{5} - \sqrt{2}) = \sqrt{2} L$
$r = \frac{\sqrt{2}}{\sqrt{5} - \sqrt{2}} L$
Since $r$ is the distance to the of point $B$ (where $-4 q$ is located), the correct option is $A$.

Two point charges $+10 q$ and $-4 q$ are located at $x=0$ and $x=L$ respectively. What is the location of a point on the $x$-axis from the origin, where the net electric field due to these two point charges is zero? $(r = \text{required distance})$

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