Electric Potential Due to Dipole

Have you ever tried filling up a glass of water to the absolute edge of the top? As you pour smaller and smaller increments of water into the glass, you'll notice that it takes a substantially large volume of water for the water to finally overflow and spill out from the glass. This is due to a phenomenon called surface tension, which allows the water at the top surface of the glass to have tension and reduce to a minimum surface area, thus resulting in a larger amount of water being held at the surface.

Fig. 1 - The surface tension of water is what allows for the water droplet to keep its shape.

The phenomenon of water tension is due to the fact that the water molecules making up the liquid are electric dipoles. As water molecules are made up of hydrogen and oxygen, where hydrogen is positively charged and oxygen is negatively charged, the two opposite charges create an electric dipole. This allows for the water molecules to form a strong bond with each other, resulting in surface tension. This article will be focusing on the electric potential due to a dipole, keep reading to find out more!

Potential Due to a Dipole at Any Point

Before calculating the electric potential due to an electric dipole at a point, it is essential to understand what is an electric potential and electric dipole individually. So let's start with an electric dipole.

The measurement of the strength of an electric dipole is given by the dipole moment \(\vec{p}\). It is a vector quantity whose direction is from negative to positive charge along the dipole length.

Let's consider two point charges \(+q\) and \(-q\) separated by a certain distance say \(2a\).

Fig. 2 - The electric dipole moment of an electric dipole points from a negative charge \(-q\) towards a positive charge \(+q\).

The magnitude of an electric dipole moment of a dipole shown in the diagram is\[\left|\vec{p}\right|=2qa,\] and its SI unit is the coulomb-meter \(\left(\mathrm{C\,m}\right)\).

Now, let's learn what an electric potential is and why it is important to know about the electric potential due to a dipole.

An electric potential determines the direction of the flow of charge. A positive electric charge always accelerates from high potential to low potential. The flow of charge stops as soon as the potential gradient becomes zero.

Imagine a single-point charge \(+q\) at a point O. The electric potential at the point which is at a distance of \(r\) from the point O is the amount of work done in moving per unit of positive charge to move it from infinity to the point P against the electrostatic force of repulsion:\[V=\frac{1}{4\pi\varepsilon_0}\frac{q}{r}.\]

From the above diagram,

1. The position vector of P from the center of the dipole is \(\vec{\text{OP}}=\vec{r}\),

2. \(\angle \text{BOP}=\theta\),

3. Distance of P from the ends of the dipole, i.e., A and B, is \(\text{AP}=r_1\) and \(\text{BP}=r_2\).

Electric Potential at point P due to a charge \(-q\) at point A is \[V_1=-\frac{q}{4\pi\epsilon_0r_1}\tag{1}\] Similarly, electric potential at point P due to a charge \(+q\) at point B is \[V_2=\frac{q}{4\pi\epsilon_0r_2}.\tag{2}\] By using the superposition principle, the electric potential at point P due to both the charges at points A and B is \[V=V_1+V_2.\] Substituting the values from equations (1) and (2) in the above equation, \[\begin{align*}V&=-\frac{q}{4\pi\epsilon_0r_1}+\frac{q}{4\pi\epsilon_0r_2}\\V&=\frac{q}{4\pi\epsilon_0}{\left(-\frac{1}{r_1}+\frac{1}{r_2}\right)}\\V&=\frac{q}{4\pi\epsilon_0}{\left(\frac{1}{r_2}-\frac{1}{r_1}\right)}.\tag{3}\end{align*}\]

This equation gives the value of electric potential due to a dipole. In the next part, we will get the formula of electric potential due to a dipole which can be used in axial (when the observation point is on the line parallel to the length of the dipole) and equatorial case(when the observation is on the line perpendicular to the length of the dipole).

Electric Potential due to a Dipole Formula

Now we have established our equation for the electric potential of a dipole as

\[ V = \frac{p}{4 \pi \epsilon_0 r^2}\cos{\left(\theta\right)},\]

where \(p\) is the electric dipole moment measured in \(\mathrm{C \, m}\), \(V\) is the electric potential measured in units of volts \(\mathrm{V}\), \(\epsilon_0\) is the permittivity of free-space with a value of \( 8.85 \times 10^{-12} \, \mathrm{\frac{F}{m}}\), and \(\theta\) is the angle between the point and the dipole measured in \(\mathrm{rads}\).

Electric Potential due to a Dipole on an Axial Line

Imagine an observation point P on the line, passing through both ends of the dipole.

Fig. 5 - Observation point P on the line passing through both the ends of a dipole such that an angle between the position vector of P from the center of a dipole and the length of the dipole is \(0^\circ\).

As the angle between OP and AB is \(0^\circ\), thus, the electric potential due to a dipole at point P, in this case, is \[\begin{align*}V_{\mathrm{axial}}&=\frac{p}{4\pi\epsilon_0r^2}\cos{0^\circ}\\V_{\mathrm{axial}}&=\frac{p}{4\pi\epsilon_0r^2}\end{align*}\] or \[V_{\mathrm{axial}}=\frac{q2a}{4\pi\epsilon_0r^2}.\] This above equation gives an electric potential due to a dipole on the axial line.

The equation shows that:

1. the electric potential due to a dipole is maximum on the axial line,

2. the electric potential varies inversely with the square of the distance from an observation point to the center of the dipole,

3. the electric potential varies directly with the length of the dipole,

4. the electric potential varies directly with the strength of the charge on either side of the dipole.

Electric Potential due to a Dipole on an Equatorial Line

Imagine an observation point P is on the line perpendicular to the length of the dipole.

Fig. 6 - Observation point P on the line perpendicular to the length of an electric dipole such that the angle between its position vector from the center of the dipole and the length of the dipole is \(90^\circ\)

From the above Figure, the angle between OP and AB is \(90^\circ\). Thus, the electric potential due to a dipole at point P, in this case, is \[\begin{align*}V_{\mathrm{eq}}&=\frac{p}{4\pi\epsilon_0r^2}\cos{90^\circ}\\V_{\mathrm{eq}}&=0\end{align*}\] The above equation shows that an electric potential due to a dipole on an equatorial line is zero. This is logical because the distance to either charge is the same on this line, so the point experiences equal but opposite voltages due to these charges.