7.3 Calculations of electric potential  (Page 3/7)

Now let us consider the special case when the distance of the point P from the dipole is much greater than the distance between the charges in the dipole, $r\gg d;$ for example, when we are interested in the electric potential due to a polarized molecule such as a water molecule. This is not so far (infinity) that we can simply treat the potential as zero, but the distance is great enough that we can simplify our calculations relative to the previous example.

We start by noting that in [link] the potential is given by

where

This is still the exact formula. To take advantage of the fact that $r\gg d,$ we rewrite the radii in terms of polar coordinates, with $x=r\text{sin}\theta$ and $z=r\text{cos}\theta$ . This gives us

We can simplify this expression by pulling r out of the root,

and then multiplying out the parentheses

The last term in the root is small enough to be negligible (remember $r\gg d,$ and hence ${\left(d\text{/}r\right)}^2$ is extremely small, effectively zero to the level we will probably be measuring), leaving us with

Using the binomial approximation (a standard result from the mathematics of series, when $\alpha$ is small)

and substituting this into our formula for $V_P$ , we get

This may be written more conveniently if we define a new quantity, the electric dipole moment    ,

where these vectors point from the negative to the positive charge. Note that this has magnitude qd . This quantity allows us to write the potential at point P due to a dipole at the origin as

A diagram of the application of this formula is shown in [link] .

There are also higher-order moments, for quadrupoles, octupoles, and so on. You will see these in future classes.

Potential of continuous charge distributions

We have been working with point charges a great deal, but what about continuous charge distributions? Recall from [link] that

We may treat a continuous charge distribution as a collection of infinitesimally separated individual points. This yields the integral

for the potential at a point P . Note that r is the distance from each individual point in the charge distribution to the point P . As we saw in Electric Charges and Fields , the infinitesimal charges are given by

where $\lambda$ is linear charge density, $\sigma$ is the charge per unit area, and $\rho$ is the charge per unit volume.

Potential of a line of charge

Find the electric potential of a uniformly charged, nonconducting wire with linear density $\lambda$ (coulomb/meter) and length L at a point that lies on a line that divides the wire into two equal parts.

Strategy

To set up the problem, we choose Cartesian coordinates in such a way as to exploit the symmetry in the problem as much as possible. We place the origin at the center of the wire and orient the y -axis along the wire so that the ends of the wire are at $y=\pm L\text{/}2$ . The field point P is in the xy -plane and since the choice of axes is up to us, we choose the x -axis to pass through the field point P , as shown in [link] .

Solution

Consider a small element of the charge distribution between y and $y+dy$ . The charge in this cell is $dq=\lambda dy$ and the distance from the cell to the field point P is $\sqrt{x^2+y^2}.$ Therefore, the potential becomes

$\begin{array}{cc}{V}_P\hfill & =k{\displaystyle \int \frac{dq}{r}=k{\displaystyle {\int }_{\text{−}L\text{/}2}^{L\text{/}2}\frac{\lambda dy}{\sqrt{x^2+y^2}}=k\lambda {\left[\text{ln}\left(y+\sqrt{y^2+x^2}\right)\right]}_{\text{−}L\text{/}2}^{L\text{/}2}}}\hfill \\ & =k\lambda \left[\text{ln}\left(\left(\frac{L}{2}\right)+\sqrt{{\left(\frac{L}{2}\right)}^2+x^2}\right)-\text{ln}\left(\left(-\frac{L}{2}\right)+\sqrt{{\left(-\frac{L}{2}\right)}^2+x^2}\right)\right]\hfill \\ & =k\lambda \text{ln}\left[\frac{L+\sqrt{L^2+4x^2}}{\text{−}L+\sqrt{L^2+4x^2}}\right].\hfill \end{array}$

Significance

Note that this was simpler than the equivalent problem for electric field, due to the use of scalar quantities. Recall that we expect the zero level of the potential to be at infinity, when we have a finite charge. To examine this, we take the limit of the above potential as x approaches infinity; in this case, the terms inside the natural log approach one, and hence the potential approaches zero in this limit. Note that we could have done this problem equivalently in cylindrical coordinates; the only effect would be to substitute r for x and z for y .

Got questions? Get instant answers now!