# Electric force and influence

Physics0001
Hi,

I'm dealing with a more or less trivial question. Let's have a look on two situations.

1. Consider a (negative) charged metal plate. If the plate is infinit in size it will produce a perfekt homogeneous electric field. If we now place a second plate parallel to the first one the electric field will influnce (positive) charge on the surface of the second plate. In an excersie they say that there is no resulting electric force attached to the second plate since the field is homogeneous and the force resulting from the field on the positive surface charge will be compensated by the force to the negative surface charge influenced on the other sied of the second plate.
2. If we rub a plastic foil on some cloths it gets also charged. According to the gaussian law the electric field should also be homogeneous. After putting it onto the wall it will stick on the wall due the influnced charged and the electric force.
What is the difference between the two situations? why do we have a force in one but not in the other situation? Has this to do with the fact that the wall is an insulator so that this is an effect of a dielectric?

Gold Member
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In case 1 of course there's a force on the 2nd plate. To keep it at a fixed distance you need to compensate this force by fixing the plate. You get the force per unit area by evaluating the Maxwell stress tensor

https://en.wikipedia.org/wiki/Maxwell_stress_tensor

• etotheipi
Physics0001
But what about the fact that the electric field of the charged plate is uniform? The force attached to every point is equal in direction and magnitude. So the resulting force should be zero?

In case 1 of course there's a force on the 2nd plate. To keep it at a fixed distance you need to compensate this force by fixing the plate. You get the force per unit area by evaluating the Maxwell stress tensor

https://en.wikipedia.org/wiki/Maxwell_stress_tensor

I'm not sure how to use this if we have a charged and an uncharged plate. Can you give more details?

Gold Member
2022 Award
Well, if you have a plate with positive charge and a negative charge, how can these plates not attract each other?

Take two infinite conducting plates parallel to the 12-plane of a Cartesian coordinate system one with positive surface charge density ##\sigma>0## at ##z=0## and one with surface charge density ##-\sigma<0##. The electric field is easily calculated by symmetry to be only within the plates, ##\vec{E}=\vec{e}_3\sigma/\epsilon_0##.

The Maxwell stress tensor is
$$T_{ij}=\epsilon_0 E_i E_j -\frac{1}{2} \epsilon_0 \vec{E}^2 \delta_{ij}$$
The force per unit on the lower plate with unit-normal vector ##\vec{n}=\vec{e}_3## is given by
$$T_i=T_{ij} n_j=\epsilon_0 E_i E_3 - \frac{1}{2} \epsilon_0 \vec{E}^2 \delta_{i3}$$
or
$$\vec{T}=\frac{1}{2 \epsilon_0} \sigma^2 \vec{e}_3.$$
As expected the force (per unit area) on the lower plate is upwards. A similar calculate for the upper plate gives the same force per unit area in the opposite direction (as it must be by symmetry).

• etotheipi
jartsa
If we rub a plastic foil on some cloths it gets also charged. According to the gaussian law the electric field should also be homogeneous. After putting it onto the wall it will stick on the wall due the influnced charged and the electric force.

So the charged foil clings to the wall. So there must be an electric field that attracts the charged foil.

Hmm well, the wall is quite thick and not infinitely tall and wide. And the wall is connected to a huge ball, the earth. So I would say that because of those reasons the charges in the wall create an electric field and the charged foil experiences a force ##E*q## in that electric field.

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Physics0001
Well, if you have a plate with positive charge and a negative charge, how can these plates not attract each other?

Take two infinite conducting plates parallel to the 12-plane of a Cartesian coordinate system one with positive surface charge density ##\sigma>0## at ##z=0## and one with surface charge density ##-\sigma<0##. The electric field is easily calculated by symmetry to be only within the plates, ##\vec{E}=\vec{e}_3\sigma/\epsilon_0##.

The Maxwell stress tensor is
$$T_{ij}=\epsilon_0 E_i E_j -\frac{1}{2} \epsilon_0 \vec{E}^2 \delta_{ij}$$
The force per unit on the lower plate with unit-normal vector ##\vec{n}=\vec{e}_3## is given by
$$T_i=T_{ij} n_j=\epsilon_0 E_i E_3 - \frac{1}{2} \epsilon_0 \vec{E}^2 \delta_{i3}$$
or
$$\vec{T}=\frac{1}{2 \epsilon_0} \sigma^2 \vec{e}_3.$$
As expected the force (per unit area) on the lower plate is upwards. A similar calculate for the upper plate gives the same force per unit area in the opposite direction (as it must be by symmetry).
But in your example both plates are charged with ##\vert \sigma>0\vert##. In my example the lower plate has a charge density ##-\sigma>0## an the upper plate has ##\sigma_0 = 0##. So the situation is asymmetric per se.

Physics0001
So the charged foil clings to the wall. So there must be an electric field that attracts the charged foil.

Hmm well, the wall is quite thick and not infinitely tall and wide. And the wall is connected to a huge ball, the earth. So I would say that because of those reasons the charges in the wall create an electric field and the charged foil experiences a force ##E*q## in that electric field.

The charge is influenced like in situation 1. But since the wall is a dielectric it depends on the polarisation which value the influenced charge density has.