Question 82 of 311

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 ______.

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Answer: 2

Let us assume at distance $y$, force is maximum.
Now, force at a distance $y$ is given by, ${F}_{net}=2F\cos \theta$
So, ${F}_{net}=\dfrac{2Kq{q}_{0}y}{{\left({y}^{2}+{a}^{2}\right)}^{\dfrac{3}{2}}}$
For ${F}_{net}$ to be maximum, $\dfrac{d{F}_{net}}{dy}=0$
Or $Kq{q}_{0}\dfrac{{\left({y}^{2}+{a}^{2}\right)}^{\dfrac{3}{2}}-y\dfrac{3}{2}\times 2y{\left({y}^{2}+{a}^{2}\right)}^{\dfrac{1}{2}}}{{\left({y}^{2}+{a}^{2}\right)}^{3}}=0$
Simplifying, we get $y=\dfrac{a}{\sqrt{2}}$.
Hence, the value of $x=2$.