# Charge above an infinite conductor

1. Mar 28, 2014

### danielakkerma

Hello everyone!
1. The problem statement, all variables and given/known data
A charge, +q, is placed above an infinite conducting slab located at z<=0, at (0,0,d). Find the potential everywhere in space, without using the image-charge method.
2. Relevant equations
Laplace's equation(and its solution in spherical coordinates).
(CGS units used throughout)
3. The attempt at a solution
Firstly, I considered that according to Laplace's equation, the potential in space should a superposition of the potential emanating from the slab, with the one generated by the point-charge. So I get:
$$V = V_c + V_q$$
Where:
$$V_c = \sum_{n=0}^{\infty} (A_n r^n + \frac{B_n}{r^{n+1}}) P_n(\cos(\theta)) \\ V_q = \frac{q}{\sqrt{r^2+d^2 -2 r d \cos(\theta)}}$$
Where Pn is the ordinary Legendre polynomial of Nth order.
Then, I expanded Vq into a series form(using Legendre's development), and separating Vc according to basic boundary conditions(V = 0, at infinity. V<∞, as r→0). I get:
V(r, \theta) = \left\lbrace \begin{align} \frac{q}{d}\sum \left( \left(\frac{r}{d}\right)^n + \frac{d}{q} A_n r^n \right) P_n(\cos(\theta)), && r < d \\ \frac{q}{r}\sum \left( \left(\frac{d}{r}\right)^n + \frac{1}{q} \frac{B_n}{r^n} \right) P_n(\cos(\theta)), && r>d \end{align} \right.
Now I apply the most important boundary cond. on the potential: at its surface, i.e. when z = 0, or, in other words, when θ=π/2, I should get an equipotential. Here, I've calibrated it, to be simply zero. I then get the following two equations(one for each of the separate values of the potential: r<d & r>d):
$$\frac{q}{d}\sum \left( \left(\frac{r}{d}\right)^n + \frac{d}{q} A_n r^n \right) P_n(0) = 0 \\ \frac{q}{r}\sum \left( \left(\frac{d}{r}\right)^n + \frac{1}{q} \frac{B_n}{r^n} \right) P_n(0) = 0$$
Now, for all n=2k+1 the terms in the sum vanish instantly(and hold for any coefficient therein), since P2k+1(0) = 0, by the anti-symmetry of odd Legendre polys.
The values of the even coefficients can then easily be determined.
But that still leaves the question of the odd ones, open.
How do I obtain them?
No reasonable guess as to their value can get me close to the expected value -- which is what one would obtain through image charges.
As always,
Would appreciate any advice or aid,
Daniel

Last edited: Mar 28, 2014
2. Mar 29, 2014

### BvU

There is another boundary condition, having to do with the conducting of the slab: the E field lines at the slab cannot have a radial component!

3. Mar 29, 2014

### danielakkerma

Unfortunately, taking the gradient of E, and projecting onto the radial direction, at θ=Pi/2 creates a problem.
It does, as you rightly point out, equal Etangential, and that in turn should be zero; however, when taking it, I obtain again(for instance, when r < d):
$$n \left (q \left (\frac{r}{d}\right )^{n-1} + A_n r^{n-1} \right ) P_n(0) = 0$$
Which leads unfortunately to an erroneous result:
That is, once again, realizing that only even components do not identically vanish, I arrive at:
$$\left (q \left (\frac{r}{d}\right )^{(2n-1)} + A_{2n} r^{(2n)-1} \right )=0 \Rightarrow A_{2n} = -\frac{q}{d^{2n-1}}$$
However, I obtain here the exact same result as from the potential. Hence, it's not getting me anywhere.
Hence, I am still stumped.
Whereto forward?
Thanks,
Daniel

Last edited: Mar 29, 2014
4. Mar 30, 2014

### danielakkerma

I hate to bump this,
But does anyone have any ideas as to how to proceed?

5. Mar 30, 2014

### BvU

The slab plane has $\theta = 0$, not $\pi/2$

6. Mar 30, 2014

### danielakkerma

I don't quite follow

Sorry,
I don't quite get that.
The plane (x-y) which is located where z=0 is attained when θ = π/2, since, in the the (standard) spherical system: $r_z = r \cos(\theta)$.
If I were to apply θ=0 on E, I would obtain the field perpendicular to the surface, which, according to Gauss' law should give the surface charge dist., but (that) which is unknown to me.
If you could therefore explain what exactly you meant by θ=0, that would be much appreciated.
Thanks again!
Daniel

7. Mar 30, 2014

### BvU

Yep, sorry, my mistake. Not good at all.
Have to re-think this one...does V = 0 at $\cos\theta = 0$ imply that $\vec E$ only has a $\theta$ component there...

8. Mar 30, 2014

### danielakkerma

At θ=0 the vector would be along the z-axis, aligned with the external point-charge. V is not zero there.
If I plug in θ=Pi/2, I would indeed see that theta is perpendicular to the surface. Still, since I don't know the induced surface charge dist., I doesn't advance me anywhere.

9. Mar 31, 2014

### BvU

[Thought I posted this yesterday-don't find it back]

Looked around a bit, came up with next to nothing. Quoting Popovic (Colorado) on subject of point charge above a large, grounded conducting sheet: "Although it's possible to determine this distribution starting from an integral equation, ..." and goes on with image charge.

At $\theta = \pi/2$ (ahem..) the radial component of field from surface charge has to cancel same from point charge at z = d. Should be feasible to work this out.

Were the spherical coordinates and the spherical harmonics in the formulation of the exercise, or did you bring them in yourself ?

10. Mar 31, 2014

### danielakkerma

It was in fact my idea to develop a series formulation of the potential $\frac{1}{|\vec{r}-\vec{r}'|}$ with Legendre's polynomials, as that would give me an orthonormal basis, which in turn should have made finding the coefficients easier. Plus, it seemed like the natural thing to do, given that the solution to Laplace's equation from the conductor could also be written through Legendre's expansion. So that was the reasoning behind that.
Viz. your previous suggestion as well, I had already derived the radial field, at θ = Pi/2, which I wrote here(Once again, only even coefficient do not vanish identically):
$$\left (q \left (\frac{r}{d}\right )^{(2n-1)} + A_{2n} r^{(2n)-1} \right )=0 \Rightarrow A_{2n} = -\frac{q}{d^{2n-1}}$$
Plugging in r=d, still gives me the same value of A2n. So that's unfortunately not leading anywhere.
I think however that the problem may be reoriented: knowing the solution in advance permits me to establish what coefficients I would need for the series to converge to the desired function; As I've said, it must be an alternating series of the form $A_n \propto (-1)^{n}$.
So perhaps that's what we should be focusing on; namely, a boundary condition that could generate such a term. That's where I am still stuck.
Thanks again!

11. Apr 2, 2014

### danielakkerma

Once again, apologize for bumping...

12. Apr 3, 2014

### BvU

Same thing again -- built a post, doesn't show up. Must have not submitted the thing. Anyway:

Radial component of E field from q at the conductor location is $$E = {1\over 4\pi\epsilon_0}{q\over r^2 + d^2}{r\over \sqrt{ r^2 + d^2}}\quad\quad {\rm (a)}$$
Has to be compensated by opposite and equal component due to surface charge distribution on conductor. Can assume circular symmetry so $\sigma(\vec r) = \sigma(r)$ in x,y plane. Radial component is $$E={1\over 4\pi\epsilon_0} \int_0^r\int_0^{2\pi} {\sigma(r') \over (r- r' \cos\phi)^2 + (r'sin\phi)^2}{r'\sin\phi \, d\phi\, dr' \over \sqrt{(r- r' \cos\phi)^2 + (r'sin\phi)^2}}$$

Setting the outcome of this integral equal to -(a) is the integral equation Popovic mentions. Then still have to work out the expression for E in all space. A lot of work, so I'm open to suggestions to do it more efficiently.

Errede 6 has a lot of useful info.

13. Apr 6, 2014

### TSny

Hello, danielakkerma.

It took me quite a while to see how to add another condition that will allow you to get the even-n coefficients. Finally thought of something that appears to work.

See if you can argue that at the surface of the conducting plane, the normal component of E from Vq must equal the normal component of E from Vc. The normal component of E at the conductor is Eθ = -(1/r)∂V/∂θ evaluated at θ = π/2.

This introduces a derivative of each Pn(cosθ) with respect to θ in the series expansions. I found the following identity to be useful:

(1-x2)P'n(x) = nPn-1(x)-nxPn(x) $\;\;\;$ (e.g., Jackson's text page 90)

Here, x = cosθ and the identity simplifies nicely for x = 0 (corresponding to θ = π/2).

Last edited: Apr 6, 2014
14. Apr 7, 2014

### TSny

Sorry, I meant to say "odd-n coefficients".

15. Apr 8, 2014

### BvU

Danny isn't happy yet, I take it ?

16. Apr 8, 2014

### danielakkerma

Thank you both!

Thanks so much for your efforts, BvU & TSny! Sorry it has taken me a while to reply!
I looked over your suggestions, BvU, and I fear that such an integral equation would be impossible to solve, without a priori working out σ -- i.e. the surface charge distribution. Furthermore, it is typically necessarily to solve for the potentials(as far as I see it) in order to actually obtain the former(σ). Hence, unless you could delve more into the subject, I don't see how one might advance with that, as is.

TSny, it too occurred to me that one ought to consider the electric field in the θ direction, as you suggest.
However, so far, I haven't found a compelling reason to justify it -- even though, as you rightly point out, it will produce the correct result.
The strongest possible rationalization of your method would be to consider the flux through an element of the surface; However, since the charge on the conductor through some small parcel dS would necessarily be smaller than the net flux from the field of the charge through the same surface(since σ on the conductor is not uniform), I don't see how one can quantify it, through that.
If you have an other ideas on how to substantiate it, I would appreciate, as always, any offers thereunto.
But I think we're now closer than ever!
Thanks again, to both of you, for your amazing help!
Daniel

17. Apr 8, 2014

### TSny

There will be a nonuniform, induced negative surface charge density on the surface of the conducting slab. At a point just outside or inside the conductor, there will be an electric field produced by all of the surface charge. This electric field will have both normal and radial components.

Because the surface is a flat plane, the field produced by the surface charge must be "mirror symmetric" with respect to the plane as shown in the left figure of the attachment. Thus the normal component just inside the conductor will be equal in magnitude but opposite in direction of the normal component just outside.

Now add in the normal component of the field due to the point charge as shown in the right figure. The normal component of this field is continuous as you pass through the surface of the conductor.

We know that the total field inside the conductor must be zero. So, that tells us something about the magnitudes of the normal components of the two fields inside the conductor. Then we can make a conclusion about the magnitudes of the normal components just outside the conductor.

#### Attached Files:

• ###### Normal components.png
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18. Apr 8, 2014

### danielakkerma

TSny,
Thank you very much! I finally have my head around the solution. You're absolutely right(as I see it, now) that the view of fields nearest to the conductor can be considered through their analogues within the slab. And furthermore, the normal components are related through the need to have the electric field identically zero within.
Once again, it's a very cogent elucidation! (and I am especially beholden for the useful illustrations you provided).
As an aside, I had calculated the radial components at θ = 0 (which would merely reproduce the normal direction) -- reached the same coefficients, and thus, made sure that the definitions thereof are consistent(as one would expect).
So, once more,
Many thanks for your help, TSny & BvU,
Daniel