 # Electric Potential Of A Dipole And System Of Charges

Before we understand the characteristics of the electric potential of a dipole, let us quickly review our understanding of dipole and electric potential. An electric potential is the amount of work needed to move a unit positive charge from a reference point to a specific point inside an electric field without producing acceleration. A dipole is a pair of opposite charges with equal magnitudes separated by a distance, d.

The electric potential due to a point charge q at a distance of r from that charge is given by,

$$\begin{array}{l}V\end{array}$$
=
$$\begin{array}{l}\frac{1}{4πε_0}~\frac{q}{r}\end{array}$$

Where
$$\begin{array}{l}ε_0\end{array}$$
is the permittivity of free space.

The electric potential is a scalar field whose gradient becomes the electrostatic vector field. Since it is a scalar field, it is easy to find the potential due to a system of charges. It is the summation of the electric potentials at a point due to individual charges.

Thus, we can write the net electric potential due to the individual potentials contributed by charges as

$$\begin{array}{l}V_{net} = \sum\limits_{i}~V_{i}\end{array}$$

$$\begin{array}{l}V_{net} = \frac{1}{4πε_0}~\sum\limits_{i}~\frac{q_i}{r_i}\end{array}$$

Let us use this concept to find the electric field of a dipole. Let the distance between the point P and the positive and negative charges be r+ and r- respectively. Let the distance from the midpoint of the dipole be r. Let this vector subtend an angle θ to the dipole axis. Thus using the above theorem we have,

$$\begin{array}{l}V\end{array}$$
=
$$\begin{array}{l}\sum\limits_{i}~V_i\end{array}$$
=
$$\begin{array}{l}V_{+}~+~V_{-}\end{array}$$

$$\begin{array}{l}V\end{array}$$
=
$$\begin{array}{l}\frac{1}{4πε_0}~\left(\frac{q}{r_{+}}~ +~\frac{-q}{r_{-}}\right)\end{array}$$ $$\begin{array}{l}V\end{array}$$
=
$$\begin{array}{l}\frac{q}{4πε_0}\left(\frac{1}{r_{+}}~ -~\frac{1}{r_{-}} \right)\end{array}$$

$$\begin{array}{l}V\end{array}$$
=
$$\begin{array}{l}\frac{q}{4πε_0}~\left(\frac{r_{-}~-~r_{+}}{r_{+}~ r_{-}}\right)\end{array}$$

If the point P is sufficiently far from the dipole, then we can approximate

$$\begin{array}{l}r_{-}\end{array}$$
$$\begin{array}{l}r_{+}\end{array}$$
$$\begin{array}{l}r\end{array}$$

And by drawing a line perpendicular to r- from +q we can write,

$$\begin{array}{l}r_{-}~-~r_{+}\end{array}$$
$$\begin{array}{l}d cos~θ\end{array}$$

Thus, we can write the potential as:

$$\begin{array}{l}V\end{array}$$
=
$$\begin{array}{l}\frac{q}{4πε_0}~\left(\frac{d ~cos~θ}{r^2}\right)\end{array}$$

We know that the magnitude of the electric dipole moment is:

$$\begin{array}{l}|\overrightarrow{p}|\end{array}$$
=
$$\begin{array}{l}p\end{array}$$
=
$$\begin{array}{l}q~.~d\end{array}$$

Thus, electric potential due to a dipole at a point far away from the dipole is given by,

$$\begin{array}{l}V\end{array}$$
=
$$\begin{array}{l}\frac{1}{4πε_0}~ \frac{p~ cos~θ}{r^2}\end{array}$$

Notice that the potential falls off by

$$\begin{array}{l}r^2\end{array}$$
while the electric field falls off by
$$\begin{array}{l}r^3\end{array}$$
. Also note that when the angle is 90 degrees, the point
$$\begin{array}{l}P\end{array}$$
is equidistant to both charges and the electric potential is zero. When θ > 90⁰, the potential is negative because the point P is closer to the negative charge.

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