Two equal positive point charges are separated by a distance 2 a . 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 \fraca\sqrtx . The value of x is \underline\hspace2cm .

Electrostatics PYQ — JEE Main 2023 · 1 Feb · Shift 1 · Q63 with Solution

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Correct answer: 2

Let the two equal positive charges be $+Q$ placed at points $(-a,0)$ and $(+a,0)$ .

A test charge $q_0$ is placed on the perpendicular bisector (equatorial line) at point $(0,y)$ .

We need the value of $y$ for which the net force on $q_0$ is maximum.

Distance of the point from each charge is

$r=\sqrt{a^2+y^2}$

Electric field due to one charge has magnitude

$E_1=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$

Because of symmetry, the horizontal components cancel and the vertical components add.

Vertical component due to one charge:

=\frac{1}{4\pi\varepsilon_0}\frac{Qy}{(a^2+y^2)^{3/2}}$$ So total electric field is $$E=2E_{1y}=\frac{1}{4\pi\varepsilon_0}\frac{2Qy}{(a^2+y^2)^{3/2}}$$ Hence force on test charge $q_0$ is $$F=q_0E=\frac{1}{4\pi\varepsilon_0}\frac{2Qq_0y}{(a^2+y^2)^{3/2}}$$ To maximize force, we maximize $$f(y)=\frac{y}{(a^2+y^2)^{3/2}}$$ --- 3. **Differentiate and set derivative to zero** $$f(y)=y(a^2+y^2)^{-3/2}$$ Differentiate: $$\frac{df}{dy}=(a^2+y^2)^{-3/2}+y\left(-\frac{3}{2}\right)(a^2+y^2)^{-5/2}(2y)$$ $$\frac{df}{dy}=(a^2+y^2)^{-5/2}\left[(a^2+y^2)-3y^2\right]$$ $$\frac{df}{dy}=(a^2+y^2)^{-5/2}(a^2-2y^2)$$ For maximum, $$a^2-2y^2=0$$ $$2y^2=a^2$$ $$y=\frac{a}{\sqrt{2}}$$ --- 4. **Compare with the given form** Given distance is $$\frac{a}{\sqrt{x}}$$ So, $$\frac{a}{\sqrt{x}}=\frac{a}{\sqrt{2}}$$ Therefore, $$x=2$$ --- 5. **Comparison with stored answer** Stored correct answer = $2$ Our derived answer = $2$ So the stored answer is correct.