Attraction force of atoms inside a metal sphere, near a charged plate

In summary, the atoms inside the small sphere feel the attraction force because there are more electrons on the sphere-surface facing the plate.
  • #1
Peter Ahlman
18
0
Hi, I have read about electric charge and tried to understand one particular thing but still couldnt:

If we positively charge a big metal plate with very high voltage and approach to it a small solid neutral metal sphere, the smaller sphere obviously will get attracted, i.e. a force will assert itself between the metal bodies.
My question: will the atoms inside of the small sphere feel the attraction force, i.e. will the atoms in the "inside-half-of-sphere" facing the plate "want to attract" to the plate, or is it only the atoms on the surface that are really attracted, and why is it so in this latter case? Gauss law sais the electric charge inside a conductor is always 0 yes, but I really don't exactly understand how that aplies to an attracted ball to a plate, there is some dissymmetry, right?

Thanks and
Regards
 
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  • #2
The net force comes from free electrons moving on the surface. You can calculate various forces for atoms and electrons inside, but they will all cancel as the electric field is 0.
 
  • #3
mfb, thank yo for the reply which is where I am again not sure if understand this very part correctly or not.

1-You said: " The net force comes from free electrons moving on the surface", I can understand that OLNY the electrons on the sphere-surface facing the plate are the ones responsible for the attraction force and thus the electrons on that surface are moving . good so far.

2-This is the part I keep reading everywhere but I don't understand, the actions/electrons inside the metal sphere asyou said:

"..they will all cancel as the electric field is 0."

I wish there is a simple drawing/diagram of a sphere and the charges that shows this, For instance I know what "canceling out means", by all means all atoms inside a "neutral" sphere do cancel out as well, but what I would love to know is(knwoing that they cancel out):

Are the atoms´ charge distribution different inside a charged/affected solid sphere if we magically see them under magnification? certainly the atoms will cancel out but is their distrubtion/electron position in an "charged/attracted" sphere just like in a neutral sphere.. really?

Thanks
Regards
 
  • #4
Well, you have missing electrons on the far side, which will give some repulsion (more positive charges there => more repulsion). In a uniform field, both effects cancel and the total force on the ball is 0.

Are the atoms´ charge distribution different inside a charged/affected solid sphere if we magically see them under magnification?
Different from a solid sphere without external fields? No, they are not.
 
  • #5
Peter Ahlman said:
Hi, I have read about electric charge and tried to understand one particular thing but still couldnt:

If we positively charge a big metal plate with very high voltage and approach to it a small solid neutral metal sphere, the smaller sphere obviously will get attracted, i.e. a force will assert itself between the metal bodies.

Where did you get that? I don't think charged plate would attract neutrally charged metal sphere any more than it would attract plastic ping-pong ball.


My question: will the atoms inside of the small sphere feel the attraction force, i.e. will the atoms in the "inside-half-of-sphere" facing the plate "want to attract" to the plate, or is it only the atoms on the surface that are really attracted, and why is it so in this latter case? Gauss law sais the electric charge inside a conductor is always 0 yes, but I really don't exactly understand how that aplies to an attracted ball to a plate, there is some dissymmetry, right?

If something is electrically charged then there is asymmetry in a way that there is either more or less electrons compared to protons over given area, but you need this asymmetry in both bodies for there to be electric force between them. Or so I would think.
 
  • #6
It is a little hard for me to describe what I want to.. (heck, I may not even know exactly what I want to know) :), anyway, If we have a charged solid sphere and I shall put it this way: (Im extremely exaggerating the case here of course)

Is the charged sphere trying to pull itself apart from the inside out, are the atoms trying to push outwards? ( ofcourse the sphere is held together still by the enormous natural bond forces and of course there will never be a net force since all the charge is pointing outwards in all directions)
 
  • #7
Peter Ahlman said:
It is a little hard for me to describe what I want to.. (heck, I may not even know exactly what I want to know) :), anyway, If we have a charged solid sphere and I shall put it this way: (Im extremely exaggerating the case here of course)

Is the charged sphere trying to pull itself apart from the inside out, are the atoms trying to push outwards? ( ofcourse the sphere is held together still by the enormous natural bond forces and of course there will never be a net force since all the charge is pointing outwards in all directions)

Electric and magnetic field strength falls off pretty rapidly with distance. At close range those forces are what define molecules arrangement and holds them together. Molecules and atoms are attracting and repealing, trying to orient and move in such way where negative charge seeks positive charge and north magnetic fields seek south magnetic fields, and vice versa. Once they find each other they "lock in", form bonds, and when opposite fields are close together like that, then macroscopically they appear neutral.

If there is equal amount of electrons and protons they will tend to arrange in such way to get close to each other and neutralize, that's why things are mostly neutrally charged in nature, macroscopically. You need external force to disturb their balance, or introduce excess of either protons or electrons, in order to make them macroscopically charged.
 
Last edited:

1. What is the attraction force between atoms inside a metal sphere and a charged plate?

The attraction force between atoms inside a metal sphere and a charged plate is known as the electrostatic force. It is caused by the difference in electric charge between the positively charged atoms in the metal sphere and the negatively charged plate.

2. How does the distance between the metal sphere and the charged plate affect the attraction force?

The attraction force between the atoms inside the metal sphere and the charged plate decreases as the distance between them increases. This is because the electrostatic force follows an inverse square law, meaning that the force decreases exponentially as the distance increases.

3. Can the attraction force between atoms inside a metal sphere and a charged plate be measured?

Yes, the attraction force between atoms inside a metal sphere and a charged plate can be measured using a variety of experimental methods, such as a force sensor or a torsion balance. These methods can accurately measure the electrostatic force between the two objects.

4. How does the charge of the plate affect the attraction force?

The attraction force between atoms inside a metal sphere and a charged plate is directly proportional to the magnitude of the charge on the plate. This means that the larger the charge on the plate, the stronger the attraction force will be between the two objects.

5. What other factors can affect the attraction force between atoms inside a metal sphere and a charged plate?

Aside from the distance and the charge, the type of metal and the shape of the metal sphere can also affect the attraction force. Different metals have different properties that can affect their electrostatic interactions, and the shape of the metal sphere can also change the distribution of charge on its surface, thus altering the attraction force with the charged plate.

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