JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2024
- Hard
- Numerical
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 \(\left.\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right)\)

Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(27\)
Step-by-step Solution
Detailed explanation
\(\mathrm{V}=\frac{\mathrm{K} \dot{\mathrm{p}} \cdot \overrightarrow{\mathrm{r}}}{\mathrm{r}^3}=\frac{9 \times 10^9(6 \mathrm{q} \ell)}{\mathrm{r}^2} \cos \left(120^{\circ}\right)\)
\(=-(27)\left(\frac{\mathrm{q} \ell}{\mathrm{r}^2}\right) \times 10^9 \mathrm{Nm}^2 \mathrm{c}^{-2}\)
\(\Rightarrow \alpha=27\)
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