Electric Dipole Potential: Axial & Equatorial Points

1. The Electric Dipole

Before we calculate potential, let's quickly define our subject. An electric dipole is a pair of equal and opposite point charges, $+q$ and $-q$, separated by a small distance, conventionally denoted as $2a$.

A crucial quantity associated with a dipole is its dipole moment ($\vec{p}$). It's a vector defined as:

where $2\vec{a}$ is the vector pointing from the negative charge $-q$ to the positive charge $+q$. The magnitude of the dipole moment is $p=q(2a)$, and its SI unit is the coulomb-meter (C·m).

2. Potential at a Point on the Axial Line

The axial line is the line passing through both charges of the dipole. Let's find the potential at a point P on this line, at a distance $r$ from the center of the dipole.

The logic is straightforward: we will calculate the potential at P due to $+q$, then the potential at P due to $-q$, and add them up.

• The distance of point P from the charge $+q$ is $(r-a)$.
• The distance of point P from the charge $-q$ is $(r+a)$.

Using the formula for potential due to a point charge, the potential at P due to $+q$ is:
$V_{+q}=\frac{1}{4\pi {ϵ}_0}\frac{q}{r-a}$
And the potential at P due to $-q$ is:
$V_{-q}=\frac{1}{4\pi {ϵ}_0}\frac{-q}{r+a}$
The total potential $V_{axial}$ is the algebraic sum:
$V_{axial}=V_{+q}+V_{-q}=\frac{q}{4\pi {ϵ}_0}\left[\frac{1}{r-a}-\frac{1}{r+a}\right]$
By taking a common denominator, we get:
$V_{axial}=\frac{q}{4\pi {ϵ}_0}\left[\frac{(r+a)-(r-a)}{(r-a)(r+a)}\right]=\frac{q}{4\pi {ϵ}_0}\left[\frac{2a}{r^2-a^2}\right]$
Recognizing that $p=q(2a)$, we can write the final expression:
$V_{axial}=\frac{1}{4\pi {ϵ}_0}\frac{p}{r^2-a^2}$

The Short Dipole Approximation

In most practical scenarios, the dipole is very small compared to the distance at which we are measuring the potential (i.e., $r\gg a$). In this case, $a^2$ is negligible compared to $r^2$. Applying this approximation:
$V_{axial}\approx \frac{1}{4\pi {ϵ}_0}\frac{p}{r^2}(\text{for }r\gg a)$
Notice how the potential from a dipole falls off as $1/r^2$, which is faster than the $1/r$ fall-off for a single point charge.

To see these derivations animated, the following video provides a very clear and concise walkthrough for both the axial and equatorial cases.

3. Potential at a Point on the Equatorial Line

The equatorial line is the line passing through the center of the dipole and perpendicular to its axis (the perpendicular bisector).

Let's find the potential at point P. Again, we use the superposition principle.
By the Pythagorean theorem, the distance of point P from both $+q$ and $-q$ is the same:
$\text{Distance}=\sqrt{r^2+a^2}$
Now, let's calculate the potentials:

• Potential at P due to $+q$: $V_{+q}=\frac{1}{4\pi {ϵ}_0}\frac{q}{\sqrt{r^2+a^2}}$
• Potential at P due to $-q$: $V_{-q}=\frac{1}{4\pi {ϵ}_0}\frac{-q}{\sqrt{r^2+a^2}}$

The total potential $V_{equatorial}$ is the sum:
$V_{equatorial}=V_{+q}+V_{-q}=\frac{1}{4\pi {ϵ}_0}\left(\frac{q}{\sqrt{r^2+a^2}}-\frac{q}{\sqrt{r^2+a^2}}\right)$
$V_{equatorial}=0$
This is a remarkable and important result: The electric potential at every point on the equatorial line of an electric dipole is zero. This happens because any point on the equatorial line is equidistant from the two equal and opposite charges, so their potentials exactly cancel out.

For a textual reference and a more detailed breakdown, you can review the "Special Cases" section in the following resource.

4. Numerical Practice

As you requested, let's put this theory into practice. The best way to solidify your understanding is by solving problems. The following resource contains several worked-out examples.

Conclusion

Today, we've extended our understanding of potential to the electric dipole. The process relied on the superposition principle you learned in the last lesson, but the results are unique to the dipole's structure.

Here are the key takeaways:

• Axial Potential: The potential on the axial line of a dipole is $V_{axial}=\frac{1}{4\pi {ϵ}_0}\frac{p}{r^2-a^2}$. For a short dipole ($r\gg a$), this simplifies to $V_{axial}\approx \frac{1}{4\pi {ϵ}_0}\frac{p}{r^2}$.
• Equatorial Potential: The potential at any point on the equatorial line is always zero.
• Distance Dependence: The potential of a dipole decreases as $1/r^2$ at large distances, faster than the $1/r$ decrease for a single point charge.

In our next lesson, we will explore the concept of equipotential surfaces. These are surfaces in space where the potential is constant. We've just found one such surface: the equatorial plane of a dipole is an equipotential surface with a potential of zero! We will see how to visualize these surfaces for various charge configurations and understand their relationship with the electric field.