How Do You Calculate Electric Potential and Field Components of a Dipole?

[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
[tex](x,y)=(Rcos(\theta)-a,Rsin(\theta))[/tex]
[tex]r=\sqrt{R^{2}-2aRcos(\theta)+a^{2}}[/tex]

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

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

Now to take the limit where r>>a I think it would end up looking like this:
[tex]V=k(\frac{Q}{\sqrt{R^{2}-2Rcos(\theta)}}+\frac{-Q}{\sqrt{R^{2}+2Rcos(\theta)}})[/tex]
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 E_r and E_θ I think all I need to do is:
[tex]\frac{\partial V}{\partial r}[/tex] and [tex]\frac{\partial V}{\partial \theta}[/tex]

Have I made any mistakes anywhere?

 

[vela]

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
[tex](x,y)=(Rcos(\theta-a),Rsin(\theta)[/tex]
[tex]r=\sqrt{R^{2}-2aRcos(\theta)+a^{2}}[/tex]

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

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.

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

Looks okay.

Now to take the limit where r>>a I think it would end up looking like this:
[tex]V=k(\frac{Q}{\sqrt{R^{2}-2Rcos(\theta)}}+\frac{-Q}{\sqrt{R^{2}+2Rcos(\theta)}})[/tex]
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.

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:
[tex]\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}[/tex]and then use the binomial expansion to expand the square root. Note that when [itex]a \ll R[/itex], you have [itex]a/R \ll 1[/itex] so the series will converge quickly.

As for finding E_r and E_θ I think all I need to do is:
[tex]\frac{\partial V}{\partial r}[/tex] and [tex]\frac{\partial V}{\partial \theta}[/tex]

Have I made any mistakes anywhere?

Look up the gradient in spherical coordinates. Note that your expression for E_θ wouldn't have the correct units for the electric field.

 

[Kurt Peek]

Hi Lancelot59,

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

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

As for your subsequent question on how to determine the field, the [itex]\theta[/itex]-component of the field is given by [tex]\frac{1}{R} \frac{\partial V}{\partial \theta}[/tex] with a factor [itex]1/R[/itex]. (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

 

[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!

 

[paranormal]

Your approach for part A looks correct. However, when taking the limit r>>a, it should be noted that the term a^2 in the denominator can also be considered negligible and can be ignored. This will simplify the expression to:
V = k(Q/r - Qcos(theta)/r)

For part B, the dipole moment magnitude is defined as p = 2aQ. So, you can rewrite the expression for V in terms of p as:
V = k(pcos(theta)/r^2)

For part C, the equipotential surfaces for a dipole would be a series of concentric circles around the dipole axis, with the potential decreasing as you move further away from the dipole.

For part D, you are correct in taking the derivative of V with respect to r to find Er. This will give you:
Er = -k(pcos(theta)/r^3)

For part E, you can use the same approach to find Etheta by taking the derivative of V with respect to theta. This will give you:
Etheta = -k(psin(theta)/r^2)

Overall, your approach looks correct, but be sure to double check your calculations and take into account any simplifications that can be made in certain limits.