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Electrostatic Potential and Capacitance — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsElectrostatic Potential and Capacitance5 Marks
Question
Figure shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on r for r/a >> 1, and contrast your results with thatdue to an electric dipole, and an electric monopole (i.e., a single charge).
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Answer

Four charges of same magnitude are placed at points X, Y, Y, and Z respectively, as shown in the following figure. A point is located at P, which is r distance away from point Y. The system of charges forms an electric quadrupole. It can be considered that the system of the electric quadrupole has three charges. Charge +q placed at point X
Charge -2q placed at point Y
Charge +q placed at point Z XY = YZ = a
YP = r
PX = r + a
PZ = r - a Electrostatic potential caused by the system of three charges at point P is given by, $\text{V}=\frac{1}{{4}\pi\in_{0}}\bigg[\frac{\text{q}}{\text{XP}}-\frac{{2}\text{q}}{\text{YP}}+\frac{\text{q}}{\text{ZP}}\bigg]$ $=\frac{1}{{4}\pi\in_{0}}\bigg[\frac{\text{q}}{\text{r}+\text{a}}-\frac{{2}\text{q}}{\text{r}}+\frac{\text{q}}{\text{r}-\text{a}}\bigg]$ $=\frac{\text{q}}{{4}\pi\in_{0}}\bigg[\frac{\text{r}(\text{r}-\text{a})-{2}(\text{r}+\text{a})(\text{r}-\text{a})+\text{r}(\text{r}+\text{a})}{\text{r}(\text{r}+\text{a})(\text{r}-\text{a})}\bigg]$ $=\frac{\text{q}}{{4}\pi\in_{0}}\bigg[\frac{\text{r}^2-\text{ra}-{2}\text{r}^2+{2}\text{a}^2+\text{r}^2+\text{ra}}{\text{r}(\text{r}^2-\text{a}^2)}\bigg]$ $=\frac{\text{q}}{{4}\pi\in_{0}}\bigg[\frac{2\text{a}^2}{\text{r}(\text{r}^2-\text{a}^2)}\bigg]$ $=\frac{{2}\text{qa}^2}{{4}\pi\in_{0}\text{r}^3\bigg({1}-\frac{\text{a}^2}{\text{r}^2}\bigg)}$ since $\frac{\text{r}}{\text{a}}>>{1}$ $\therefore\ \frac{\text{r}}{\text{a}}>>{1}$$\frac{\text{r}^2}{\text{a}^2}$ is taken as negligible.
$\therefore\ \text{V}=\frac{{2}\text{qa}^2}{{4}\pi\in_{0}\text{r}^3}$ It can be inferred that potential, $\text{V}\propto\frac{1}{\text{r}^3}$ However, it is known that for a dipole, $\text{V}\propto\frac{1}{\text{r}^2}$ And, for a monopole, $\text{V}\propto\frac{1}{\text{r}}$

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