A conducting shell kept in a uniform electric field

[TSny]

This is the solution for the axisymmetric case where the potential Φ is not a function of the azimuth angle φ. But here you can't ignore φ; Φ is a function of r, θ and φ.

If the direction θ = 0 is chosen parallel to the applied external field, then Φ will be independent of φ.

(It can for example easily be shown by symmetry argument that the potential is zero everywhere along a line extending from infinity to the center of the shell with the line being perpendicular to the E field).

Yes. The plane θ = $\pi$/2 is an equipotential surface where Φ = 0.

 

[TSny]

Solution to lapalace equation in spherical coordinates:

$$\Phi=\sum_{i=0}^{\infty}A_ir^i+B_ir^{-i-1}P_i(\cos\theta)$$

The $P_i(\cos\theta)$ factor should also multiply the $A_i$ term.

Your analysis is good. If $A_1$ is the only nonzero $A_i$ term, what is the only nonzero $B_i$ term?

 

[Titan97]

@TSny, $$A_1=-E_0r$$
$$B_1=-E_0a^3$$
For all $r$,
$$\Phi=-E_0\left(r-\frac{a^3}{r^2}\right)$$
This is the potential due to the applied electric field and that due to the induced charge on the shell.

Am I correct If I say that $-E_0r$ is due to applied electric field and $E_0\frac{r^3}{a^2}$ is due to the induced charge?

To get $\sigma$, should I consider the potential due to induced charge only or the net potential $\Phi$?

 

[rude man]

If the direction θ = 0 is chosen parallel to the applied external field, then Φ will be independent of φ.
Yes. The plane θ = $\pi$/2 is an equipotential surface where Φ = 0.

Yes, I failed to see that. Thanks for joining in!

 

[Titan97]

@rude man will I get to learn legendre polynomial in engineering? Or is it a part of mathematics? I just learned vector calculus because of interest and to solve this problem (I read Purcell and Morin's book).

 

[haruspex]

I did not understand the derivation. It's beyond my high school syllabus).

You can arrive at this distribution by a thought experiment. Since the induced charges must neutralise the external field within the sphere, the field they generate must be uniform within the sphere. So if we project the charges onto the plane disc boundary between them it would produce a uniform charge on the each side of the disc. Projecting that back, we find the distributions follow a cosine law.

 

[TSny]

@TSny, $$A_1=-E_0r$$
$$B_1=-E_0a^3$$
For all $r$,
$$\Phi=-E_0\left(r-\frac{a^3}{r^2}\right)$$

Good, but don't forget the contribution from $P_1(\cos \theta)$.

Am I correct If I say that $-E_0r$ is due to applied electric field and $E_0\frac{r^3}{a^2}$ is due to the induced charge?

Yes, once you fix it.

To get $\sigma$, should I consider the potential due to induced charge only or the net potential $\Phi$?

The net potential is what matters. How do you plan to get $\sigma$ from $\Phi$?

 

[rude man]

@rude man will I get to learn legendre polynomial in engineering? Or is it a part of mathematics? I just learned vector calculus because of interest and to solve this problem (I read Purcell and Morin's book).

Legendre polynomials is something you are likely to encounter in an advanced undergraduate physics (electricity & magnetism) or math course. You will most probably not see it in an introductory, engineering-level physics course; and I would say in math not before a 5th semester course. Of course, math courses vary widely in material covered, depending on the school and on prior (high school) exposure.

As TSny pointed out, the potential expression is a function of r and θ only, and in fact I find that (based on the image dipole method) it's a pretty simple expression. And the surface charge density is a function of θ only, if that will help you to your answer.

 

[Titan97]

@TSny $$E=\frac{\partial\Phi}{\partial r}\hat{r}+\frac{1}{r}\frac{\partial\Phi}{\partial\theta}\hat{\theta}$$
$$E=-E_0\left(1+\frac{2a^3}{r^3}\right)\cos\theta\hat{r}+E_0\left(r-\frac{a^3}{r^2}\right)\sin\theta\hat{\theta}$$
$$\sigma=\epsilon_0E_{r=a}=-3\epsilon_0E_0\cos\theta$$
As @rude man pointed out, charge density depends only on $\theta$.

Should I now integrate $EdA\sigma$?

 

[TSny]

@TSny $$E=\frac{\partial\Phi}{\partial r}\hat{r}+\frac{1}{r}\frac{\partial\Phi}{\partial\theta}\hat{\theta}$$
$$E=-E_0\left(1+\frac{2a^3}{r^3}\right)\cos\theta\hat{r}+E_0\left(r-\frac{a^3}{r^2}\right)\sin\theta\hat{\theta}$$
$$\sigma=\epsilon_0E_{r=a}=-3\epsilon_0E_0\cos\theta$$

OK, but don't forget the negative sign in the basic equation $\vec{E} = -\nabla V$.

Should I now integrate $EdA\sigma$?

There's a factor of 1/2 that's easy to overlook here. For example, see this video. Be sure you're awake at around time 7:45 in the video.

 

[TSny]

(I read Purcell and Morin's book).

The 1/2 factor mentioned in my previous post is derived in section 1.14 of Purcell and Morin. See equation 1.50.

 

[Titan97]

@TSny should I add the force due to external electric field and the force due to the shell on itself?

 

[TSny]

@TSny should I add the force due to external electric field and the force due to the shell on itself?

No, the formula at 7:45 in the video (or equations 1.44 and 1.50 in Purcell) gives you the total force per unit area: $$\frac{dF}{dA} = \frac{\sigma ^2}{2 \varepsilon_0}$$.

 

[Titan97]

At the surface, $E=3E_0\cos\theta$. This electric field is due to both the applied electric field and due to the induced charge.

Near a conducting surface (that is, at A) , the electric field due to the conductor is $\frac{\sigma}{\epsilon_0}$

Let E1 be field due to the charge inside the pillbox and E2 be other electric field.

Total electric field at A is $$\frac{\sigma}{\epsilon_0}=E_1+E_2$$
Inside the conductor, at surface B, $E_1=E_2$
$$E_2=\frac{\sigma}{2\epsilon_0}$$

Force is $\frac{\sigma}{2\epsilon_0}dq$

 

[TSny]

At the surface, $E=3E_0\cos\theta$. This electric field is due to both the applied electric field and due to the induced charge.

Yes.

Near a conducting surface (that is, at A) , the electric field due to the conductor is $\frac{\sigma}{\epsilon_0}$

$\sigma / \varepsilon_0$ equals the net electric field just outside the surface of the conductor. This net electric field includes the field of all charges on the conductor as well as any other charges not on the conductor and any applied external field.

 

[rude man]

You've done extremely well! Especially for a high school student!
The notion of electrostatic pressure due to surface charge density is perhaps a bit mysterious. Here is a link giving a good derivation of the pressure formula at a conductor of p = σ²/2ε₀

http://www.physicspages.com/2011/10/31/electrostatic-pressure/

 

[Titan97]

$$dF=\frac{\sigma}{2\epsilon_0}\sigma dA=\frac{9}{2}\epsilon_0E_0^2\cos^2\theta\hat{r}$$
$$dA=2\pi a^2\cos\theta\sin\theta d\theta$$
Only force along $z$ remains (that is, along rcosθ). The other components cancels out.
$$dF=9\epsilon_0E_0^2\pi a^2\cos^3\theta\sin\theta d\theta$$
$$F=\frac{9}{4}\epsilon_0E_0^2\pi a^2$$

 

[TSny]

$$F=\frac{9}{4}\epsilon_0E_0^2\pi a^2$$

I think that's correct. Great work!

 

[rude man]

$$dF=\frac{\sigma}{2\epsilon_0}\sigma dA=\frac{9}{2}\epsilon_0E_0^2\cos^2\theta\hat{r}$$
$$dA=2\pi a^2\cos\theta\sin\theta d\theta$$

I have dA = a² sinθ dθ dφ = 2πa²sinθ dθ.

 

[Titan97]

The radius of the ring is $a\sin\theta$. Its width is $ad\theta$. Yes. The area is $2\pi a^2\sin\theta d\theta$.

Then $$dF=\frac{\sigma}{2\epsilon_0}\sigma dA\cos\theta$$
since other component cancels out.

The final answer is same. I think I made some typing error. (I also missed 'dA' in the first line).

 

[rude man]

The radius of the ring is $a\sin\theta$. Its width is $ad\theta$. Yes. The area is $2\pi a^2\sin\theta d\theta$.

Then $$dF=\frac{\sigma}{2\epsilon_0}\sigma dA\cos\theta$$
.

Why the cos θ term? Looks to me like it should be sin θ, not cos θ, since the "pull" is maximum at the equator where θ = π/2, and zero at the poles where θ = 0 and π.

So that would make $$dF=\frac{\sigma}{2\epsilon_0}\sigma dA\sin\theta$$, with
dA = 2πa²sin θ dθ.

 

[Titan97]

@rude man

I need to take the cosine component.

 

[rude man]

@rude man
View attachment 99332
I need to take the cosine component.

You're quite right. I had the wrong orientation of the shell w/r/t my coordinate system.