Question

Four particles (one of charge q,one of charge 3q,and two of charge -2q)are placed as shown in Fig. 3.31, each a distance from the origin. Find a

simple approximate formula for the potential, valid at points far from the origin.

(Express your answer in spherical coordinates.)

Short answer

The potential at the origin for the point charge is $V=\frac{1}{2\pi {\epsilon }_0}\frac{2qa\cos \theta }{r^2}$.

Step-by-step solution

Define function

The dipole moment is the product of charge and distance of separation charge from the origin.

Write the expression for dipole moment.

$\mathit{p}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{1}}^{\mathbf{n}}{\mathit{q}}_{\mathbf{i}}{\mathit{r}}_{\mathbf{i}}$ ……. (1)

Here, is the charge of $i^{th}$particle,$r_i$ is the distance of separation charge form the origin.

Given data

The position of particle is shown in figure.

It consist of charges as q ,3q-2q,-2q. The separation between the charges denoted by a.

Determine potential

Write the expression for the total dipole moment due to four charges.

$\begin{array}{l}P=\left(3qa\hat{z}-qa\hat{z}\right)+\left(-2qa\hat{y}+-2qa\left(-\hat{y}\right)\right)\\ =2qa\hat{z}+\left(2qa\hat{y}\right)+2qa\hat{y}.........\left(2\right)\\ =2qa\hat{z}\end{array}$

Write the expression for the position of the point in spherical coordinates.

$$\begin{gathered} x=r\sin \theta \cos \theta \\ q=r\sin \theta \sin \theta \\ z=r\cos \theta \end{gathered}$$

The scalar product of dipole P and position vector $\hat{r}$is,

$P.\hat{r}$ ......(3)

Substitute $2qa\hat{z}$ for P in equation (3).

$$\begin{gathered} P.\hat{r}=2qa\hat{z}.\hat{r} \\ =2qa\left(\hat{z}.\hat{r}\right) \\ =2qa\left(\hat{r}\cos \theta .\hat{r}\right) \end{gathered}$$

Substitute $\hat{r}.\hat{r}=1$

Thus,

$P.\hat{r}=2qa\cos \theta .......\left(4\right)$

Write the equation for potential at distance rdue to point charge q.

$V=\frac{1}{4\pi {\epsilon }_0}\frac{P.\hat{r}}{r^2}......\left(5\right)$

Substitute 2qacos$\theta$ for P .$\hat{r}$, in equation (5)

$V=\frac{1}{4\pi {\epsilon }_0}\frac{2qa\cos \theta }{r^2}$

Thus, the potential at the origin for the point charge is $V=\frac{1}{4\pi \epsilon }\frac{2qa\cos \theta }{r^2}$.