- #1

pepos04

- 41

- 0

- 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.

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]**
Last edited: