Point charge with very thin metal sheet along a spherical surface

[pepos04]

Homework Statement

Consider a point charge $q$ placed at a point $(0,0, b \geq 0)$ on the positive half-space of the $z$ (horizontal) in the presence of a very thin isolated and overall neutral metal sheet arranged along half of the spherical surface of radius $a$ centered in the origin and placed in the negative half-space of the $z$. Discuss qualitatively the force to which the $q$ charge is subjected and determine whether or not for some positive value of $b$ it can be repulsive, that is, directed as $+\vec k$ the versor of the $z$ axis. Consider in particular the $b \ll a$ case and the $0\leq b \leq a$ case.

Relevant Equations

Maybe $\frac{q^2}{4 \pi \epsilon_0} (\frac{a}{b} -\frac{b}{a} )=0$

This is what I thought. The electrostatic force between a point charge and a metal plate is not simply given by Coulomb’s law, because the metal plate is not a point charge, but a conductor that can redistribute its charges in response to the external field. A useful method to calculate the electrostatic force in this situation could be the method of image charges, which replaces the metal plate with an imaginary point charge of opposite sign at a symmetric position with respect to the plate, but I'm not sure about that. This way, the problem could be reduced to finding the force between two point charges in free space.

But, in this case, the metal plate is not flat, but curved along a spherical surface. This makes the problem more complicated, because the image charge would be not simply located at the mirror image of the original charge, but would depend on the curvature of the plate. IMO, there is a special case where the image charge method can be applied easily: when the metal plate is a hemisphere, and the original charge is on the axis of symmetry. In this case, the image charge could be located at the center of the sphere, and has a magnitude equal to the original charge multiplied by the ratio of the radius to the distance from the center.

In general, the electrostatic force between a point charge and a metal plate is always attractive, because the plate induces an opposite charge on its surface that is closer to the original charge than the rest of the plate. However, in the special case of a hemispherical plate and an axial charge, there is a possibility of a repulsive force, depending on the position of the charge. This is because the image charge at the center can be larger than the original charge, and thus dominate the force.

To find the condition for a repulsive force, we could equate the magnitude of the force to zero and solve for the distance of the charge from the center. This gives me: $$\frac{q^2}{4 \pi \epsilon_0} (\frac{a}{b} -\frac{b}{a} )=0$$ Solving for $b$, we get two solutions: $b=0$ and $b=a$. The first solution corresponds to the charge being at the center of the sphere, where the force is zero by symmetry. The second solution corresponds to the charge being on the surface of the hemisphere, where the force is also zero because the image charge cancels out the original charge.

Therefore, for any positive value of $b$ between $0$ and $a$, the force is attractive, and for any positive value of $b$ greater than $a$, the force is repulsive. In particular, when $b \gg a$, the force is approximately given by Coulomb’s law, because the image charge is very small and far away. When $0 \leq b \leq a$, the force is smaller than Coulomb’s law, because the image charge partially cancels out the original charge. I don't know if this reasoning can work (almost certainly not, which is why I would like your help), but this is all I have been able to produce. Both qualitative and quantitative descriptions are needed.

[Mentor Note: Post edited for legibility]

 

[BvU]

Homework Statement: Consider a point charge $q$ placed at a point $(0,0, b \geq 0)$ on the positive half-space of the $z$ (horizontal) in the presence of a very thin isolated and overall neutral metal sheet arranged along half of the spherical surface of radius $a$ centered in the origin and placed in the negative half-space of the $z$. ...

The first solution corresponds to the charge being at the center of the sphere, where the force is zero by symmetry.

Hi,

I read in the problem statement that there is only a half spherical surface...

$\$

 

[haruspex]

when the metal plate is a hemisphere, and the original charge is on the axis of symmetry. In this case, the image charge could be located at the center of the sphere, and has a magnitude equal to the original charge multiplied by the ratio of the radius to the distance from the center.

I do not understand how you arrive at that relationship, or why the image would be at the centre of curvature.

To find the condition for a repulsive force, we could equate the magnitude of the force to zero and solve for the distance of the charge from the center. This gives me: $$\frac{q^2}{4 \pi \epsilon_0} (\frac{a}{b} -\frac{b}{a} )=0$$

At b=0 that gives infinity, not zero. As you say, it should give zero.

Don’t expect to be able to do much in the way of actual equations. Just see if you can come up with an argument that for b>0 it will be attracted to the hemisphere.

 

[haruspex]

Hi,

I read in the problem statement that there is only a half spherical surface...

$\$

The OP seems to understand that, and used "center of the sphere" to mean centre of curvature.

 

[pepos04]

@BvUYouare right. Any advice on how to approach this? Also, I copied wrong: you have to analyze the case $b \gg a$, not $b \ll a$. Sorry.

 

[haruspex]

@BvUYouare right. Any advice on how to approach this? Also, I copied wrong: you have to analyze the case $b \gg a$, not $b \ll a$. Sorry.

In the b>>a case, can you think of a simple charge arrangement that might approximate that on the hemisphere?

 

[pepos04]

@haruspex I am not sure whether we want to work with the hemisphere. Maybe it is better with a whole sphere or a flat mirror, through method of images, or not? In fact, the field of the charge would polarise the hemisphere and produce a complicated field that is mostly dipole, then the dipole would attract the charge. Not well. I had also thought that, if z=0 outside of the hemisphere is also a flat metallic sheet, maybe there is a simple solution. As it is, it might be a complicated sum. The fact is that I cannot see how to work with all that. Any hint is appreciated.

 

[BvU]

I had also thought that, if z=0 outside of the hemisphere is also a flat metallic sheet

I do not read that in the problem statement.

For the case $b>>a$ I would ask myself : does the shape of the sheet matter very much ? And combine that with the given that it's neutral.

$\$

 

[haruspex]

mostly dipole

Yes.

Maybe it is better with a whole sphere or a flat mirror, through method of images

The method of images works with grounded conductors, either whole spheres or (a special case of that) infinite flat sheets.
It occurred to me that an isolated spherical conductor can be equated to two charges: first, the usual mirror charge q' for the grounded sphere, plus a second charge q" at the sphere's centre. The trouble is, I do not see how to find the relationship between q" and the charge on the sphere, Q.

As I wrote, it seems you are not expected to solve it algebraically. The author is looking for more of a handwaving argument. And I suspect the looked-for answer is that it will always be attracted.

I had also thought that, if z=0 outside of the hemisphere is also a flat metallic sheet, maybe there is a simple solution.

Not sure what you are suggesting. You are adding a large metallic sheet to the point charge+hemisphere arrangement ?

 

[pepos04]

@haruspex The problem is light years away from being easy and bland. Of course, the obvious answer would be 'force is always attractive', but this is not the case. I was told that there are two separate cases to be analysed because the two responses are different: repulsive in one case, attractive in another. I am beginning to think that even the method of image charges is no longer sufficient, even though- as you say- an analytic solution is not really needed. Incidentally I asked the author of the problem and he said he would give me the analytical answer today, but it is complicated. So I would like your help in trying to understand it...

 

[haruspex]

I can make a handwaving argument for why it might be repulsive at small b>0.
With b=0, as we know, there is no induced charge distribution on the hemisphere. As the charge, +ve say, is moved away from z=0, -ve charges will be attracted to the part of the hemisphere near z=0 and the distal part will become positively charged.
Although the negative charges are slightly nearer the point charge, they form a ring almost around it, so exert little net field in the z direction. The positive charges pushed towards z=-a, on the other hand, act fully in the z direction.
I do not consider this proof, but it is suggestive. Also, it is very unclear at what value of b it would return to no net force.

 

[pepos04]

@haruspex Great argument! Below is mine, but I assume it is incorrect… The negative and positive charges induced by q in the half-shell are in the details indeterminable. What is certain is that the charge-q' closest to q must be in absolute value equal to the charge induced +q' furthest away. I denote by $0\leq\theta\leq\pi/2$ the angle formed with the radius a perpendicular to z by a radius of the half circle section with the sheet. The area with a negative induced charge is between 0 and $\theta_0$ and the area with a positive induced charge is between $\theta_0$ and $\pi/2$. The induced charges in the half-shell are indeterminable in detail. What I can say is that by denoting with $\sigma^{-}, A^{-}$ and $\sigma ^{+}A^{+}$ charge density and area of competence their product $qq^{+}$ and $q q^{-}$ must be equal in absolute value. Now suppose we divide the surface of the half-shell into infinitesimal circular crowns each $a d\theta$ wide and $a \cos\theta$ radius. The attractive forces directed as $-\vec k$ with $-\vec k$ direction of the z-axis will be by symmetry $F_{-}=q^{-} q \int_0^{\theta_0}{\frac{2} a \cos\theta} }a d\theta}{(b+a \sin\theta)^2}} and$F_{+}$obtained with the positive density and integrating between$\theta_0$and$\pi/2$. An important observation must now be made about the vertical and horizontal components of these forces. If we consider two diametrically opposed crown elements, the components of the force exerted on q elide while the horizontal components along the z-axis double. Only the direct horizontal components count as$-\vec k$if they are attractive and as$-\vec k$if they are repulsive. Thus the attractive-repulsive difference represents the total force acting on q. In detail, however, unexpected results may occur as the text suggests a) b>>a. The distance that determines the intensity of the force can be considered in both cases as$b^2$but the repulsive horizontal components i.e. directed as$\vec k$may for some b prevail, because as can be seen they may be greater than the attractive ones that 'end before' the others and with relevant vertical components that elide. Repulsive components, on the other hand, have smaller vertical components and which decrease until they cancel out at$\pi/2$. Hence there may exist values of b for which the total resultant is directed as$\vec k$. (b) 0<b<a . When$b\longrightarrow 0$, practically the attractive force is all vertical and elides while the repulsive one retains an appreciable maximum horizontal component at$\pi/2$where the vertical one disappears. Again, there may be values of b that render the total force as direct as$\vec k##. What is wrong with this reasoning and my assumptions? Please adjust the shot where possible.

 

[haruspex]

Your posts would be a lot more readable if you were to make use of paragraphs. There is also a problem with your LaTeX. I've had a go at fixing both, but I may have misguessed:

@haruspex Great argument! Below is mine, but I assume it is incorrect…
The negative and positive charges induced by q in the half-shell are in the details indeterminable. What is certain is that the charge-q' closest to q must be in absolute value equal to the charge induced +q' furthest away.
I denote by $0\leq\theta\leq\pi/2$ the angle formed with the radius a perpendicular to z by a radius of the half circle section with the sheet. The area with a negative induced charge is between 0 and $\theta_0$ and the area with a positive induced charge is between $\theta_0$ and $\pi/2$.

The induced charges in the half-shell are indeterminable in detail. What I can say is that by denoting with $\sigma^{-}, A^{-}$ and $\sigma ^{+}A^{+}$ charge density and area of competence their product $qq^{+}$ and $q q^{-}$ must be equal in absolute value.

Now suppose we divide the surface of the half-shell into infinitesimal circular crowns each $a d\theta$ wide and $a \cos\theta$ radius. The attractive forces directed as $-\vec k$ with $-\vec k$ direction of the z-axis will be by symmetry $F_{-}=q^{-} q \int_0^{\theta_0}\frac{2} {a \cos\theta} \frac{a d\theta}{(b+a \sin\theta)^2}$ and $F_{+}$ obtained with the positive density and integrating between $\theta_0$ and $\pi/2$.

An important observation must now be made about the vertical and horizontal components of these forces. If we consider two diametrically opposed crown elements, the components of the force exerted on q elide while the horizontal components along the z-axis double. Only the direct horizontal components count as $-\vec k$ if they are attractive and as $-\vec k$ if they are repulsive. Thus the attractive-repulsive difference represents the total force acting on q. In detail, however, unexpected results may occur as the text suggests

a) b>>a. The distance that determines the intensity of the force can be considered in both cases as $b^2$ but the repulsive horizontal components i.e. directed as $\vec k$ may for some b prevail, because as can be seen they may be greater than the attractive ones that 'end before' the others and with relevant vertical components that elide. Repulsive components, on the other hand, have smaller vertical components and which decrease until they cancel out at $\pi/2$. Hence there may exist values of b for which the total resultant is directed as $\vec k$.

(b) 0<b<a . When $b\longrightarrow 0$, practically the attractive force is all vertical and elides while the repulsive one retains an appreciable maximum horizontal component at $\pi/2$ where the vertical one disappears. Again, there may be values of b that render the total force as direct as $\vec k$.
What is wrong with this reasoning and my assumptions? Please adjust the shot where possible.

 

[pepos04]

@haruspex Yes, that's right. So, what is wrong with this reasoning and my assumptions? Please adjust the shot where possible. https://arxiv.org/abs/1007.2175 This is the solution, but I cannot understand it. Why is a conformal map from the outside of a semicircle to the outside of a metal disc of radius $\frac{R}{\sqrt 2}$? Why does a logarithmic function appear in the method of image charges? However, I want to see if my solution will work...

 

[haruspex]

https://arxiv.org/abs/1007.2175 This is the solution, but I cannot understand it.

I love the energy graph argument, very neat.

Why is a conformal map from the outside of a semicircle to the outside of a metal disc of radius $\frac{R}{\sqrt 2}$?

I do not understand what you are asking. It is like "why is a banana?"
Are you asking why it is useful, or how it achieves that mapping, or ….?

I'll look more at your solution later.

 

[pepos04]

@haruspex Yes, sorry, I misplaced the question. I didn't understand how that conformal map is obtained, nor why the radius of the disc has to be the one indicated by the paper (perhaps it has to do with the area of the semicircle), nor why the method of image charges returns that result with logarithms.

 

[haruspex]

@haruspex Yes, sorry, I misplaced the question. I didn't understand how that conformal map is obtained, nor why the radius of the disc has to be the one indicated by the paper (perhaps it has to do with the area of the semicircle), nor why the method of image charges returns that result with logarithms.

Hmmm.. I cannot make that mapping do what is claimed.
Taking R=1, and u to be a point on the semicircle, $u=\cos(\theta)+i\sin(\theta)=c+is$, I get $2h=(c-sc)+i(1+s+s\sqrt{1+s^2})$.

Plotting that for theta from π to 2π I see a curve that is not quite circular in appearance and doesn't close at the top.
Indeed, if we plug u=R, u=-R into the map we get $R\frac{1+i}2$, $R\frac{-1+i}2$, which matches my plot.

 

[pepos04]

@haruspex Let me know when you get that conformal map from the text: yours looks very similar to me! Instead, I had thought of applying Mobius transformations and other isomorphisms... Also, have you looked at my solution? What's wrong with it?

 

[pepos04]

@haruspex How did you get $2h=(c-sc)+i(1+s+s\sqrt{1+s^2})$. Could you explain?

 

[haruspex]

@haruspex How did you get $2h=(c-sc)+i(1+s+s\sqrt{1+s^2})$. Could you explain?

I substituted $u=c+is, R=1$ in the equation in the article.

 

[pepos04]

@haruspex Excuse me, can you tell me where the condition $\phi_y(x) + q \mathbf{x}| = 0$ on page 4 of the article comes from? How is this boundary condition obtained? The conditions before and after this one are supposed to be Neumann and Dirichlet conditions, but I didn't understand what this one is. Could you clarify it for me? Also, I did not understand why it is said that $\frac{q}{2} \nabla_x \phi_x(x) =- F(x)$. Could you let me know this other condition?

 

[haruspex]

@haruspex Excuse me, can you tell me where the condition $\phi_y(x) + q \mathbf{x}| = 0$ on page 4 of the article comes from? How is this boundary condition obtained? The conditions before and after this one are supposed to be Neumann and Dirichlet conditions, but I didn't understand what this one is. Could you clarify it for me?

Do you mean "$\phi_y(x)+q\ln|x|=0$ for $x\rightarrow\infty$"?
I interpret that as meaning that at large distances from the origin the potential approximates $q\ln|x|$. That follows from the field approximating $q/|x|$, which it would in two dimensions.

Also, I did not understand why it is said that $\frac{q}{2} \nabla_x \phi_x(x) =- F(x)$. Could you let me know this other condition?

Did you understand $\bar\phi$? This is the potential due to the induced charge distribution. It differs from the net potential, $\phi$, by the potential due to the point charge alone, $-q\ln|x|$.
I'm struggling with $\bar\phi_x(y)=\bar\phi_y(x)$. If I am interpreting the notation correctly, $\bar\phi_y(x)$ is the potential at x due to the induced charges when the point charge is at y. Why should that be the same as the potential at y due to the induced charges when the point charge is at x?