Electrostatics Question 18
Question 18 - 2024 (31 Jan Shift 2)
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 $\mathrm{P}$ at a distance $\mathrm{r}$ from centre $\mathrm{O}$ is $-\alpha\left[\frac{q l}{r^{2}}\right] \times 10^{9} \mathrm{~V}$, where the value of $\alpha$ is . (Use $\left.\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}\right)$
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Answer: (27)
$\mathrm{V}=\frac{\mathrm{K} \overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{r}}}{\mathrm{r}^{3}}=\frac{9 \times 10^{9}(6 \mathrm{q} \ell)}{\mathrm{r}^{2}} \cos \left(120^{\circ}\right)$
$=-(27)\left(\frac{\mathrm{q} \ell}{\mathrm{r}^{2}}\right) \times 10^{9} \mathrm{Nm}^{2} \mathrm{c}^{-2}$
$\Rightarrow \alpha=27$
