The distance between charges $+\mathrm{q}$ and $-\mathrm{q}$ is $2 l$ and between $+2 \mathrm{q}$ and $-2 \mathrm{q}$ is $4 l$. The electrostatic potential at point $P$ at a distance $r$ from centre $O$ is $-\alpha\left[\frac{q l}{r^2}\right] \times 10^9 \mathrm{~V}$, where the value of $\alpha$ is____. (Use $\left.\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right)$
JEE MAIN 2024, Diffcult
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$\mathrm{V}=\frac{\mathrm{K} \dot{\mathrm{p}} \cdot \overrightarrow{\mathrm{r}}}{\mathrm{r}^3}=\frac{9 \times 10^9(6 \mathrm{q} \ell)}{\mathrm{r}^2} \cos \left(120^{\circ}\right)$
$=-(27)\left(\frac{\mathrm{q} \ell}{\mathrm{r}^2}\right) \times 10^9 \mathrm{Nm}^2 \mathrm{c}^{-2}$
$\Rightarrow \alpha=27$
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