# Electron between parallel plates

1. Jun 25, 2015

### Scheuerf

If you place an electron between oppositely charged parallel plates, is it true the the force on it is the same regardless of how far it is from each plate? If so how?

2. Jun 25, 2015

### Noctisdark

No, let be the potential between the two plates, we know that F = -∇U, where U is the potential energy and U = E*q*d, but E varies with inverse square of the distance ie the whole quantity vary with inverse the distance so if the electron is very far, -∇U will be very small and so the electron will feel less force, another way the think of it is that the electron is far force the electric field source so F will be small,

Last edited: Jun 25, 2015
3. Jun 25, 2015

### abbas_majidi

I think F is constant, because E is constant between two plates with different potential. Same plates with different charges have not equal potentials.

4. Jun 25, 2015

### Scheuerf

In the image I attached, there is a proton between two parallel plates. Let's say the charge of the plates and distance between the plates result in a force of 2N on the proton. If F is constant that means no matter how close the proton becomes to the the negative plate the force from the negative plate alone has to be less than 2N. That doesn't make any sense to me, I would think that as the proton becomes closer to the negative plate the force would increase exponentially resulting in a much greater force than 2N

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5. Jun 25, 2015

### ZapperZ

Staff Emeritus
Why should it?

If you consider the "negative plate" as attracting, then you should also consider the "positive plate" as pushing. It may be closer, and get attracted more to the negative plate, but the positive plate is also getting farther away and pushing LESS.

This, btw, is not a good way of looking at it. For example, consider an infinite plane of charge. This will also result in a uniform E-field and has the same effect.

Zz.

6. Jun 25, 2015

### Scheuerf

I understand that the positive plate pushes less as the proton gets farther away but it will always be pushing rather than pulling. That means if F is constant the pull from the negative plate has to be less than 2N. As the distance between the negative plate and the proton approaches 0, I would think that the force on the proton would approach infinity rather than always being less than 2.

7. Jun 25, 2015

### ZapperZ

Staff Emeritus
I don't get this. Are you saying that there is a difference between these two forces? That "pushing" somehow doesn't count as much as pulling?

If so, you need to understand what a net force is. The net force is the TOTAL SUM of all the forces. It doesn't matter if it is push or pull.

Zz.

8. Jun 25, 2015

### Scheuerf

All that I meant to say was that the force from the positive and negative plates are acting in the same direction so they add up. Because the forces add up to 2N, the force that the negative plate has on the proton must be less than 2N. If it were 2N or greater, then the net force would be greater than 2N and the force on the proton would therefore not be constant at every point between the two plates.

9. Jun 25, 2015

### Staff: Mentor

The force on the proton from the piece of the plate that is directly underneath it does increase as the proton comes closer to the plate. However, the net force from the pieces of the plate that are off to the sides decreases. To see this, consider two pieces of the plate at equal distances from the center, on opposite sides of the proton. As the proton comes closer to the plate, the forces from these two pieces become more nearly opposite in direction. Their horizontal components cancel, and their vertical components become smaller and smaller.

It turns out that as the proton comes closer to the plate, the sum of the forces from all the off-center pieces decreases rapidly enough to counteract the increase in the force from the central piece of the plate, so the grand total force remains constant. To prove this, you have to use calculus.

Added to clarify: My diagram does not show two plates, but rather two situations with one plate, with the charge located at different distances. For an idealized charged plate that extends to infinity in all directions, the electric field (and force exerted on a charge) is uniform with distance. Two infinite, oppositely-charged plates exert equal forces (in the same direction of course) on a charge between them, no matter where the charge is located.

Last edited: Jun 25, 2015
10. Jun 25, 2015

### ZapperZ

Staff Emeritus
The forces from BOTH plates add up to give you that force. If the charge is closer to the positive plate, then the repulsive force will be larger than the attractive force, but still give you the same magnitude. No different than your example.

Are you still saying that you do not understand the conceptual picture of this situation?

Zz.

11. Jun 25, 2015

### Scheuerf

I understand that as the proton changes positions, the net force on it remains the same but the force from each individual plate on the proton changes. If we want to find the repulsive force between two electrons, we can use the formula F = (kq1q2)/r^2. If r can be any positive number, that means there is no limit to how strong this repulsive force can be. I would think that with the plates and proton it would be no different, but it seems that if the net force on the proton is constant at any point between the two plates, then the force from a given plate can not be higher then that constant net force and that confuses me.

12. Jun 25, 2015

### Staff: Mentor

No. First consider a single plate alone. In the idealized case of an infinite plate, the magnitude of the electric field (and force on a proton or other charge) is the same, regardless of the distance from the plate. $$F = qE = \frac{q \sigma}{2 \epsilon_0}$$ where $\sigma$ is the charge density on the plate. For two oppositely-charged plates, in the space between them, double this.

Most students usually see this first proved using Gauss's Law, in a calculus-based university-level intro physics course:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html

but it can also be proven from Coulomb's law, using calculus. The force exerted by each small piece of the plate is given by Coulomb's law, using the charge on that piece and the distance from the proton to that piece. You add the forces from all the pieces by integrating.

Coulomb's law works only for point charges, or charges small enough that you can consider them point-like for practical purposes. The plates discussed here do not meet that condition, although small pieces of them do.

Last edited: Jun 25, 2015
13. Jun 25, 2015

### Scheuerf

Thanks, I think I'm beginning to understand it now. Sorry it took me a while, my math isn't quite good enough to understand some of these concepts.

14. Jun 26, 2015

### stedwards

There's simple way to get a rough idea of what's going on without too much math.

For a charge between two closely spaced plates, the direction of the field is constrained to one dimension. $F=kq$. $k$ is some constant.

For a charge acted upon by point charge, it's just the Coulomb force. $F=kq/r^2$.

Where the field lines are constrained to two dimension the force is inverse $r$. $F=kq/r$.

The force is dimensionally depended. $F=kq/r^{n-1}$, where $n$ is the number of dimensions in which the field can spread out.

15. Sep 5, 2015

### sophiecentaur

Consider how the field between two point charges varies. See this link The field is stronger where the lines are closer together and it is lower, in the space between, where they spread out. If you put a string of + and - charges, side by side, you could imagine that the lines in between will get squeezed together by the adjacent ones. Put enough point charges on either side and the lines will end up parallel, all the way across - the field will be the same everywhere (except at the edge, where the lines will bulge out).
No Maths, just some geometry and a bit of arm waving. (That's just a demonstration, btw, and not a proof.)

16. Sep 6, 2015

### vanhees71

Maxwell's equations answer all questions. Here we have an electrostatic problem. So everything is described in terms of the potential. The finite-size plate problem is pretty tough. So usually one makes an approximation, using two infinite plates. One may be in the $xy$ plane at $z=0$ and the other parallel at $z=d$. Then you put a given potential difference (voltage) $U$ on them (e.g., by connecting a battery to them). For symmetry reasons $\Phi(\vec{x})=\Phi(z)$, and we have to solve
$$\Delta \Phi=0, \quad \Phi(z=0)=0, \quad \Phi(z=d)=U.$$
Since $\Phi(\vec{x})=\Phi(z)$ the Laplace equation simplifies to
$$\Phi''(z)=0 \; \Rightarrow \; \Phi(z)=C_1 z+ C_2,$$
and from the boundary conditions we get
$$\Phi(z=0)=0 \; \Rightarrow \; C=2=0, \quad \Phi(z=d)=0 \;\Rightarrow \; C_1=\frac{U}{d}.$$
All this is valid between the plates; outside you have $\Phi=\text{const}$.

The electric field is given by
$$\vec{E}=-\vec{\nabla} \Phi= \begin{cases} -\frac{U}{d} \vec{e}_z & \text{for} \quad 0<z<d,\\ 0 & \text{elsewhere}. \end{cases}$$
On the inner surface of the plates you have a charge-surface density given by the jump of the electric field. Assuming vacuum between the plate, i.e., $\epsilon-1$ we find from Gauß's Law
$$\sigma(z=0)=-\frac{U}{d}, \quad \sigma(z=d)=+\frac{U}{d}.$$
The force on a test particle within the plates is thus
$$\vec{F}=q \vec{E}=\text{const}.$$
All this is valid for finite plates in regions not too close to the edges and for $d \ll L$, where $L$ is a typical extension of the plates.