JEE MAIN - Physics (2024 - 31st January Evening Shift - No. 28)
The distance between charges $$+q$$ and $$-q$$ is $$2 l$$ and between $$+2 q$$ and $$-2 q$$ is $$4 l$$. The electrostatic potential at point $$P$$ at a distance $$r$$ from center $$O$$ is $$-\alpha\left[\frac{q l}{r^2}\right] \times 10^9 \mathrm{~V}$$, where the value of $$\alpha$$ is __________. (Use $$\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~Nm}^2 \mathrm{C}^{-2}$$)
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Answer
27
Explanation
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$$\begin{aligned} & \mathrm{V}=\frac{\mathrm{K} \overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{r}}}{\mathrm{r}^3}=\frac{9 \times 10^9(6 \mathrm{q} \ell)}{\mathrm{r}^2} \cos \left(120^{\circ}\right) \\\\ & =-(27)\left(\frac{\mathrm{q} \ell}{\mathrm{r}^2}\right) \times 10^9 \mathrm{Nm}^2 \mathrm{c}^{-2} \\\\ & \Rightarrow \alpha=27 \end{aligned}$$
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