Question 718: Electrostats & Capacitance

Question

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 $$-\alpha\left[\frac{q l}{r^2}ight] 	imes 10^9 \mathrm{~V}$$, where the value of $$\alpha$$ is __________. (Use $$\frac{1}{4 \pi \varepsilon_0}=9 	imes 10^9 \mathrm{~Nm}^2 \mathrm{C}^{-2}$$)

Accepted Answer

$$\begin{aligned}
& \mathrm{V}=\frac{\mathrm{K} \overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{r}}}{\mathrm{r}^3}=\frac{9 	imes 10^9(6 \mathrm{q} \ell)}{\mathrm{r}^2} \cos \left(120^{\circ}ight) \\
& =-(27)\left(\frac{\mathrm{q} \ell}{\mathrm{r}^2}ight) 	imes 10^9 \mathrm{Nm}^2 \mathrm{c}^{-2} \\
& \Rightarrow \alpha=27
\end{aligned}$$

Electrostats & Capacitance

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