# Homework Help: Electric Potential Problem

1. Oct 8, 2011

### Lancelot59

Given the following problem:

An electric dipole has charge +Q at position a and charge –Q at position -*a along the z axis.

a. Calculate the electric potential at an arbitrary point. Choose the arbitrary constant so at V=0 at infinity. Express the result in terms of r and θ (radius and polar angle with respect to z axis in spherical coordinates).

b. Simplify your result in the limit r>>a, being sure to keep the leading non-*trivial term. Write it in terms of the dipole moment magnitude p =2aQ.

c. Draw the equipotential surfaces associated with the dipole.

d. Determine the radial component Er of the electric field by taking the appropriate derivative of the potential. Check that your result makes sense.

e. Bonus: find the theta component Eθ of the electric field from the potential.

I started with A.
For Q
$$(x,y)=(Rcos(\theta)-a,Rsin(\theta))$$
$$r=\sqrt{R^{2}-2aRcos(\theta)+a^{2}}$$

For -Q
$$(x,y)=(Rcos(\theta+a),Rsin(\theta)$$
$$r=\sqrt{R^{2}+2aRcos(\theta)+a^{2}}$$

In total:
$$V=k(\frac{Q}{\sqrt{R^{2}-2aRcos(\theta)+a^{2}}}+\frac{-Q}{\sqrt{R^{2}+2aRcos(\theta)+a^{2}}})$$
I think this is correct.

Now to take the limit where r>>a I think it would end up looking like this:
$$V=k(\frac{Q}{\sqrt{R^{2}-2Rcos(\theta)}}+\frac{-Q}{\sqrt{R^{2}+2Rcos(\theta)}})$$
Since the contribution of a is negligible at that point. I do not however understand what they mean by expressing it in terms of the dipole moment magnitude.

As for finding Er and Eθ I think all I need to do is:
$$\frac{\partial V}{\partial r}$$ and $$\frac{\partial V}{\partial \theta}$$

Have I made any mistakes anywhere?

Last edited: Oct 8, 2011
2. Oct 8, 2011

### vela

Staff Emeritus
Your work is kind of sloppy. First, you should proofread what you post and fix obvious mistakes. Second, be consistent with the notation. Don't use R to stand for what the problem statement refers to as r and then use r for something else. Finally, the problem gave you a coordinate system to use. Your (x,y) notation is confusing at best and incorrect as worst.
Looks okay.
No, that's not correct. You can't simply erase all instances of a. For one thing, your expression no longer makes sense. You can't add a term with units of length squared to a term with units of length. Second, you can't simply erase factors because doing so isn't mathematically valid.

What you want to do is this instead:
$$\frac{1}{\sqrt{R^2-2aR\cos(\theta)+a^2}} = \left\{R^2\left[1-2\left(\frac{a}{R}\right)\cos\theta + \left(\frac{a}{R}\right)^2\right]\right\}^{-1/2} = \frac{1}{R}\left\{1-\left(\frac{a}{R}\right)\left[2\cos\theta + \left(\frac{a}{R}\right)\right]\right\}^{-1/2}$$and then use the binomial expansion to expand the square root. Note that when $a \ll R$, you have $a/R \ll 1$ so the series will converge quickly.
Look up the gradient in spherical coordinates. Note that your expression for Eθ wouldn't have the correct units for the electric field.

Last edited: Oct 8, 2011
3. Oct 8, 2011

### Kurt Peek

Hi Lancelot59,

Although some of your intermediate expressions don't make sense (you can't have an expression like "$\theta+a$" because $\theta$ is in radians and $a$ is in meters), I do concur with your expression for $V(r,\theta)$. (In the notation I'm used to, your constant $k=1/(4 \pi \epsilon_0)$, where $\epsilon_0$ is the permittivity of free space). Subsequently neglecting the $a^2$ terms in the square roots is indeed the first step in the approximation, although you seem to have forgotten a factor $a$ in the process.

The next step is to apply a binomial approximation. Write $$\frac{1}{\sqrt{R^2\mp 2 R a \cos(\theta)}}=\frac{1}{R}\left(1 \mp \frac{2 a \cos(\theta)}{R}\right)^{-1/2} \approx \frac{1}{R}\left(1 \pm \frac{a \cos(\theta)}{R}\right),$$ where in the last step I have used the approximation $$(1+x)^n \approx 1+n x,\qquad x \ll 1.$$ Inserting this into your equation leads to $$V(r) \approx \frac{k p \cos(\theta)}{R^2},$$ where $p$ is the dipole moment you defined in the question.

As for your subsequent question on how to determine the field, the $\theta$-component of the field is given by $$\frac{1}{R} \frac{\partial V}{\partial \theta}$$ with a factor $1/R$. (Again, it can be seen that without that factor the units would not make sense). This has to do with the fact that the gradient in spherical coordinates is not obtained by the same procedure as the gradient in Cartesian coordinates.

Finally, I would recommend you to check out H.D. Griffiths, Introduction to Electrodynamics , Example 3.10. Hope this helps!

Cheers,
Kurt

Last edited: Oct 8, 2011
4. Oct 8, 2011

### Lancelot59

Sorry, that wasn't supposed to read theta-a. I misplaced the parenthesis. I'll go over this again, and post again if I have any more confusion. Thanks!