Homework Help: Electric field

1. Oct 26, 2012

Dustinsfl

I feel like I need to incorporate $\mathbf{E} = E\hat{\mathbf{z}}$ but I don't know what to do with it.

A grounded conducting sphere of radius $a$ is placed in an (effectively) infinite uniform electric field $\mathbf{E} = E\hat{\mathbf{z}}$. The potential for a uniform electric field in the $z$-direction is given by $Er\cos\theta$. The boundary condition at the surface of the grounded sphere is that
$$u(a,\theta) = 0.$$

Use a perturbation scheme for the total potential
$$u(r,\theta) = Er\cos\theta + u'$$
to solve for the perturbation potential $u'$.

Using our giving condition, we have $u(a,\theta) = Ea\cos\theta + u'(a,\theta) = 0$. That is,
$$u'(a,\theta) = -Ea\cos\theta = \sum_{n = 0}^{\infty}\frac{A_n}{a^{n + 1}}P_n(\cos\theta).$$
From our previous work, we know that we only need the $n = 1$ term. Therefore, $-Ea\cos\theta = \frac{A_1}{a^2} P_1(\cos\theta)\Rightarrow -Ea^3 = A_1$. We have that the perturbation potential is
$$u'(r,\theta) = -Ea^3\frac{\cos\theta}{r^2}$$
and that the total potential is
$$u(r,\theta) = Er\cos\theta\left(1 - \frac{a^3}{r^3}\right).$$
$P_n$ is the nth Legendre polynomial.

2. Oct 26, 2012

aralbrec

E=Ez is in the potential function Er cos theta.

3. Oct 26, 2012

Dustinsfl

So everything is good to go then?

Thanks.

4. Oct 27, 2012

aralbrec

Yeah I think so. Compute E from E=-∇V to check. Your boundary conditions are V=0 on the sphere and E=Ez at infinity.

5. Oct 27, 2012

Dustinsfl

$$\vec{E} = -E\cos\theta\left(1 + \frac{2a^3}{r^3}\right)\hat{e_r} + E\sin\theta\left(1 - \frac{a^3}{r^3}\right)\hat{e_{\theta}}.$$

When r=a, I don't get 0, and at infinity, I have $-E\cos\theta\hat{e_r} + E\sin\theta\hat{e_{\theta}}$

6. Oct 27, 2012

aralbrec

It's important to keep the coordinate system straight because it is not quite aligned in a standard way.

Your potential solution:

u(r,θ) = Er cos θ (1 - a3 / r3)

is zero on the conducting sphere (at r=a, any θ) so that part checks out.

For your calculation of E=-∇V:

E =−Ecosθ(1+2a3/r3)er + Esinθ(1−a3/r3)

you have to be very careful with the coordinate system, which was chosen when the potential function was decided.

Remember E = Ek with potential Er cos θ

This is valid if you draw on a piece of paper the yz plane with z on the vertical axis, y on the horizontal axis and x coming out of the paper. The E field is constant and directed upward, so if you calculate potential from the origin, you only have to integrate along the z axis. Measure θ clockwise from the z axis, then the potential at coordinate (r,θ) is Ercosθ

So in your field equation, er is in the yz plane and θ is measured clockwise from the vertical z axis.

So, your final vector E field is not using standard er and unit vectors. I would convert them back to cartesian so everyone is on the same page.

Find what er and are in terms of j,k (unit vectors along the y and z axes), substitute into your E vector equation to get cartesian coordinates and then verify the field is what it should be at infinity.

Edit: sorry I've been looking at this problem piecemeal, coming back to it once in a while. The last step will have to be recognition of the symmetry to generalize the solution outside the yz plane. I hope that's right, it's been ages since I did this stuff which is part of the reason I am here -- please check to make sure it all makes sense.

Edit2: sorry dustin, one more time, this time with correct advice. The coordinate system is spherical, θ is the angle from the z axis, r is r. Potential is consistent in 3d in spherical coordinates so ∇V can be computed in spherical coordinates to get the 3d solution directly. I was misled in my haste into thinking in 2d and seeing what looked like cylindrical coordinates. Change E to cartesian or change E=Ek to spherical to see if the solution is correct at large distance.

Last edited: Oct 27, 2012
7. Oct 27, 2012

Dustinsfl

I am sort of confused since I am math not physics. What do you mean?

8. Oct 27, 2012

aralbrec

Dustin, my bad.. I did not pay close enough attention to what you were doing. The only part of my post that is accurate is "edit2"; I left the rest in there because I saw you were reading it before I edited.

Let's try again. When you said your E field might not match up I took a closer look at how your coordinate system was defined.

You start with E=Ez and arrive at a potential V=Er cos θ for that field. It sounds like you are unsure about this idea because you weren't aware that all the information about the E field is encoded in that potential function.

The reason is, in the electrostatic situation, ∇×E = 0 which means E can be uniquely determined from a potential function and E = -∇V. The potential function is a path integral V = -∫E.dl along any path (an irrotational field means the path doesn't matter). That is the review for a math guy.

To arrive at the potential function, there was a path integral from the origin to point (r,θ). Since E only has a component along the z axis, only the distance along the z axis matters in the path and the distance travelled along the z axis to the point (r,θ) is r cos θ. Since the magnitude of E is constant, the potential function for the field is V=Er cos θ.

Notice that r is distance from the origin and θ is angle from the z axis rotated clockwise. I was looking at this in the yz plane but it generalizes fine to three dimensions using spherical coordinates. The potential function is the same in 3d, r is compatible with the spherical coordinate system r, θ is the angle from the z-axis, compatible with spherical coords. This you clearly already knew so this was more for me when you said there might be a problem.

The boundary conditions we had to check was V=0 on the sphere and E=Ez at infinity.

Your solution has potential function:

u(r,θ) = Er cos θ (1 - a3 / r3)

which is everywhere zero on the sphere (r=a) so that checks out.

To check the field at large distances, calculate E=-∇V to which you came up with:

E =−Ecosθ(1+2a3/r3)er + Esinθ(1−a3/r3)

You can see the E field becomes independent of r fairly quickly as you move away from the sphere.

Next convert to cartesian by noting z = cosθ er - sinθ

If you factor the above and set r large, you'll see you end up with E=Ez

Solutions to ∇2V = 0 for the region between the sphere and infinity are unique. The boundary conditions are V=0 at r=a and E=Ez at infinity (I justify using E field as boundary conditions by looking at Green's function for electrostatics, which specifies a boundary in terms of V and ∂V/∂n (E)). The part where you say you only need n=1 term I don't know about -- if not true you only have part of the solution.

Edit: there is a problem with the sign of the potential function and that is why the direction of E doesn't line up. V = negative integral of E along the path.

Last edited: Oct 27, 2012
9. Oct 27, 2012

Dustinsfl

Thanks I understand now. I can see that everything checks outs. Do you know how I can plot the E field in Mathematica? There is a vector plot option, but when I followed the instructions in find selected function, I didn't get anything remotely correct.

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