JEE Main 2025 — Electrostatics Question with Solution

$\begin{array}{rl}& \frac{\mathrm{kq}}{(\mathrm{r}-\mathrm{a}{)}^2}=\frac{\mathrm{kq}}{(\mathrm{r}+\mathrm{a}{)}^2}+\frac{2\mathrm{kq}}{\left({\mathrm{r}}^2+{\mathrm{a}}^2\right)}\cos \theta \\ & \frac{1}{(\mathrm{r}-\mathrm{a}{)}^2}=\frac{1}{(\mathrm{r}+\mathrm{a}{)}^2}+\frac{2\mathrm{a}}{{\left({\mathrm{r}}^2+{\mathrm{a}}^2\right)}^{\frac{3}{2}}}\\ & \frac{1}{(\mathrm{r}-\mathrm{a}{)}^2}-\frac{1}{(\mathrm{r}+\mathrm{a}{)}^2}=\frac{2\mathrm{a}}{{\left({\mathrm{r}}^2+{\mathrm{a}}^2\right)}^{\frac{3}{2}}}\\ & \frac{4\mathrm{ra}}{{\left({\mathrm{r}}^2-{\mathrm{a}}^2\right)}^2}=\frac{2\mathrm{a}}{{\left({\mathrm{r}}^2+{\mathrm{a}}^2\right)}^{\frac{3}{2}}}\end{array}$

$\begin{array}{rl}& \Rightarrow \frac{2r}{{\left(r^2-a^2\right)}^2}=\frac{1}{{\left(r^2+a^2\right)}^{\frac{3}{2}}}\\ & \frac{4r^2}{{\left(r^2-a^2\right)}^4}=\frac{1}{{\left(r^2+a^2\right)}^3}\\ & \Rightarrow 4r^2{\left(r^2+a^2\right)}^3={\left(r^2-a^2\right)}^4\\ & 4r^8{\left(1+\frac{a^2}{r^2}\right)}^3=r^8{\left(1-\frac{a^2}{r^2}\right)}^4\\ & 4{\left(1+\frac{a^2}{r^2}\right)}^3={\left(1-\frac{a^2}{r^2}\right)}^4\end{array}$

Exact value cannot be solved in exam for this equation to be true

$$\left|\frac{\mathrm{a}}{\mathrm{r}}\right|>1\Rightarrow \mathrm{a}>\mathrm{r}$$

But point charge $Q$ lies between charges of dipole

1 hence electric field cannot be zero.

There for it should be bonus.

But by solving from mathematical software we are getting $\mathrm{a}/\mathrm{r}\approx 3$.