Q 12-01-091JEE MainJEE Main 2023 (1 Feb, Shift 1)Medium
Two equal positive point charges are separated by a distance $2a$. The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which force experienced by a test charge $q_0$ becomes maximum is $\dfrac{a}{\sqrt x}$. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 2
On the bisector $E=\dfrac{2kqy}{(a^2+y^2)^{3/2}}$. Setting $\dfrac{dE}{dy}=0$: $(a^2+y^2)-3y^2=0\Rightarrow y=\dfrac{a}{\sqrt2}$. So $x=2$.
Solution by Sreeraj P, M.Sc Physics