# 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,
$$V$$ = $$\frac{1}{4πε_0}~\frac{q}{r}$$

Where $$ε_0$$ 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

$$V_{net} = \sum\limits_{i}~V_{i}$$

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

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,

$$V$$ = $$\sum\limits_{i}~V_i$$ = $$V_{+}~+~V_{-}$$

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

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

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

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

$$r_{-}$$$$r_{+}$$$$r$$

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

$$r_{-}~-~r_{+}$$$$d cos~θ$$

Thus, we can write the potential as:

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

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

$$|\overrightarrow{p}|$$ = $$p$$ = $$q~.~d$$

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

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

Notice that the potential falls off by $$r^2$$ while the electric field falls off by $$r^3$$. Also note that when the angle is 90 degrees, the point $$P$$ 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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