Electrostatics Question 18
Question 18 - 2024 (31 Jan Shift 2)
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 centre $O$ is $-\alpha\left[\frac{q l}{r^{2}}\right] \times 10^{9} V$, where the value of $\alpha$ is . (Use $\left.\frac{1}{4 \pi \varepsilon _0}=9 \times 10^{9} Nm^{2} C^{-2}\right)$
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Answer: (27)
$V=\frac{K \overrightarrow{p} \cdot \overrightarrow{r}}{r^{3}}=\frac{9 \times 10^{9}(6 q \ell)}{r^{2}} \cos \left(120^{\circ}\right)$
$=-(27)\left(\frac{q \ell}{r^{2}}\right) \times 10^{9} Nm^{2} c^{-2}$
$\Rightarrow \alpha=27$
