Question

The electric potential at a point due to an electric dipole along its axis will be
A. $k\dfrac{\overrightarrow{P}\centerdot \overrightarrow{r}}{{{r}^{3}}}$
B. $k\dfrac{\overrightarrow{P}\centerdot \overrightarrow{r}}{{{r}^{2}}}$
C. $k\dfrac{\overrightarrow{P}\times \overrightarrow{r}}{{{r}^{3}}}$
D. $k\dfrac{\overrightarrow{P}\times \overrightarrow{r}}{{{r}^{2}}}$

Answer 1

Hint: You could find the expression in a very simple way by using the basics of electrostatics. You could consider the dipole as two point charges and then find the potential at the given point due to individual charge and then take the sum. Also, substitute for the dipole moment and hence get the answer.

Formula used:
Dipole moment,
$\overrightarrow{P}=2aq$

Complete step by step solution:
In the question, we are asked to find the electric field at a point due to an electric dipole along it axis among the given options.

Let ${{V}_{1}}$ be the electric potential at A due to +q and let ${{V}_{2}}$ be the electric potential at A due to –q.
We know that the electric potential at some point at distance d due to some charge q is given by,
$V=\dfrac{kq}{d}$
Now, the potential due to the given charges will be,
${{V}_{1}}=\dfrac{kq}{r+a}$
${{V}_{2}}=-\dfrac{kq}{r-a}$
Now the net potential at point A will be,
$V={{V}_{1}}+{{V}_{2}}$
$\Rightarrow V=\dfrac{kq}{r+a}-\dfrac{kq}{r-a}=kq\left( \dfrac{1}{r+a}-\dfrac{1}{r-a} \right)$
$\Rightarrow V=\dfrac{kq\left( -2a \right)}{{{r}^{2}}\left( 1-\dfrac{{{a}^{2}}}{{{r}^{2}}} \right)}$ …………………………………………….. (1)
But for$r\gg a$ ,
$\dfrac{{{a}^{2}}}{{{r}^{2}}}\approx 0$
Equation (1) can now be written as,
$V=\dfrac{kq\left( -2a \right)}{{{r}^{2}}}$
But we know that,
$\overrightarrow{P}=2aq$
So,
$V=\dfrac{k\left( -P \right)}{{{r}^{2}}}$
Now in the figure we could clearly see that the vectors $\overrightarrow{r}$ and $\overrightarrow{P}$ are directed in the opposite directions. So, we could rewrite the above equation as,
$V=\dfrac{k\overrightarrow{P}\centerdot \overrightarrow{r}}{{{r}^{3}}}$
Therefore, we found the electric potential at a point due to an electric dipole along its axis to be,
$V=\dfrac{k\overrightarrow{P}\centerdot \overrightarrow{r}}{{{r}^{3}}}$

So, the correct answer is “Option A”.

Note: We know that the unit vector that gives the direction is given by,
$\widehat{r}=\dfrac{\overrightarrow{r}}{r}$
Using this we have done the final step modification in the given solution. We could similarly derive the electric potential at an equatorial point by summing the potential due to –q and +q separately.