# What causes an electric force in one situation but not in another?

• Physics0001
In summary: In your example both plates are charged with ##\vert \sigma>0\vert##. In my example the lower plate has a charge density ##-\sigma>0## and the upper plate has ##\sigma_0 = 0##. So the situation is asymmetric per...In summary, 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.
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?

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
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?

vanhees71 said:
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?

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
Physics0001 said:
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.

Last edited:
vanhees71 said:
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.

jartsa said:
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.

## 1. How does electric force work?

Electric force is a fundamental force of nature that describes the attraction or repulsion between charged particles. It is caused by the interaction of electric fields, which are created by charged particles, and can be calculated using Coulomb's Law.

## 2. What factors affect the strength of electric force?

The strength of electric force is affected by the magnitude of the charges involved and the distance between them. The force increases as the charges get larger and decreases as the distance between them increases.

## 3. How does electric force differ from gravitational force?

Electric force is caused by the interaction of electric charges, while gravitational force is caused by the interaction of masses. Electric force can be both attractive and repulsive, while gravitational force is always attractive. Additionally, electric force is much stronger than gravitational force.

## 4. What is the role of electric force in everyday life?

Electric force plays a crucial role in many everyday objects and devices, such as electronics, appliances, and power systems. It is also responsible for phenomena like static electricity and lightning.

## 5. How is electric force related to electric fields?

Electric force is a result of the interaction between electric fields and charged particles. The strength and direction of the electric force on a charged particle can be determined by the electric field at that point. Electric fields are also used to visualize and calculate the distribution of electric force in a given space.

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