Charge distribution on a conductor.

[integration]

Homework Statement

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Here is a problem I made up, in order to try to understand electrostatics of conductors better. Suppose two large square parallel conducting plates, are in space. One has a charge of Q, and another has a charge of -q. They are put close together, from a separation of L, to a separation of l (l < L). (abs(Q) different of abs(q))

1. Describe both the inital charge distribution of both plates, and, the electric field between and on the outer sides of both plates, once equilibrium is reached.

2. After waiting, you "inject" an extra amount of charge m to the plate with inital charge -q. What happens to the charge distribution? and to the electric field?

It would be really nice if you guys could provide explanations for your answers, as I am still trying to understand electrostatics.

Homework Equations

Gauss law, Coulomb's law, E=kq/R2

The Attempt at a Solution

My idea of solving this is setting E=0 inside the conductor. Considering charges can only be on the surface, I set up surface charge densities so that the E field contribution of both plates end up cancelling each other within each plates. My difficulty lies in the fact that how do I know if my distribution is unique? And also, how do I find the possible polarization of the plates due to the other plate?


Thanks so much.

 

[anathema]

Dear forum post author,

Firstly, I would like to commend you for taking the initiative to create your own problem in order to better understand electrostatics. This is a great way to deepen your understanding of a subject.

To answer your first question, when the two plates are initially separated by a distance L, the charge distribution on each plate will be uniform. This means that the charge will be evenly distributed across the surface of each plate. The electric field between the plates will also be uniform, pointing from the positively charged plate towards the negatively charged plate. On the outer sides of the plates, the electric field will also be uniform, but in the opposite direction, pointing away from the plates.

When equilibrium is reached, the charges on each plate will redistribute themselves in such a way that the electric field inside the conductors is zero. This means that the charges will move towards the inner surface of the plates, leaving the outer surface with no net charge. This redistribution of charges will result in the electric field between the plates becoming stronger, while the electric field on the outer sides of the plates will become weaker.

Regarding your second question, when you inject an extra amount of charge m to the plate with initial charge -q, the charge distribution on that plate will change. This is because the added charge will repel the existing charges on the plate, causing them to redistribute themselves. This will result in an increase in the electric field between the plates and on the outer sides of the plate with the extra charge.

To determine the exact charge distribution and electric field in this scenario, you can use Coulomb's law and Gauss's law. Coulomb's law allows you to calculate the force between two charges, while Gauss's law allows you to calculate the electric field due to a charge distribution. By using these equations and considering the symmetry of the system, you can determine the charge distribution and electric field for each scenario.

I hope this helps in your understanding of electrostatics. Keep up the good work in creating your own problems and seeking a deeper understanding of the subject. Best of luck in your studies.

 

[nlightner]

The initial charge distribution on the two plates can be described using Coulomb's law and Gauss's law. Since the plates are parallel, the electric field between them will be uniform and directed from the positively charged plate towards the negatively charged plate. The electric field outside the plates will be zero, as the charges on the surface of the plates will cancel out the electric field produced by the opposite charges on the other plate. The charge distribution on the plates will be such that the electric field inside the conductor is zero, as you have correctly stated. This means that the surface charge density on the plates will be highest at the edges, and decrease towards the center of the plates.

When an extra charge m is injected to the plate with initial charge -q, the charge distribution on both plates will change. The extra charge will cause the electric field between the plates to become stronger, as the positive and negative charges will now be closer together. This will lead to an increase in the surface charge density on both plates, with the negatively charged plate having a higher surface charge density due to the extra charge m. This will also lead to a change in the electric field outside the plates, as the charges on the surface of the plates will now be stronger.

To determine the exact charge distribution on the plates, you will need to use Gauss's law and Coulomb's law to calculate the electric field at different points on the plates. This will give you an idea of how the charges are distributed on the plates, and whether your initial assumption of setting E=0 inside the conductor is valid. It is important to note that the charge distribution on the plates will depend on the distance between them, as well as the amount of extra charge m injected.

In terms of polarization, the plates will become polarized due to the presence of the other plate. This means that the charges on the surface of the plates will shift towards the opposite side, in order to cancel out the electric field produced by the other plate. This polarization effect will be stronger when the plates are closer together, as the electric field between them will be stronger.

I hope this explanation helps you understand electrostatics better. It is important to keep in mind that the charge distribution on a conductor is not always unique and can vary depending on the conditions and parameters of the system.