# Question about distribution of electric charges on two parallel plates

• integration
In summary: The uniqueness of the charge distribution can be determined by checking if the electric field is zero inside the conductor, and by considering the symmetry of the system.
integration
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.

PS: 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.

1. Initially, the charge distribution on each plate is uniform and equal in density to the magnitude of the charge (Q or -q) divided by the area of the plate. The electric field between the two plates is equal and opposite to the fields on the outer sides of the plates, and of magnitude Q/L^2 (the magnitude of the charge divided by the distance between the plates squared). After the two plates reach equilibrium, the electric field between them is zero. 2. When you inject an extra amount of charge m to the plate with initial charge -q, the charge distribution on that plate becomes non-uniform such that the extra charge m is located at a single point on the plate. The electric field between the two plates then increases due to the increased charge separation, and is of magnitude (Q + m)/l^2 (the sum of the charges divided by the distance between the plates squared). Additionally, the electric field on the outer side of the plate with the extra charge m increases, while the field on the outer side of the other plate decreases. The polarization of the plates due to the other plate's field can be calculated by finding the electric dipole moment of one plate due to the other plate's charge distribution, and multiplying this by the electric field of the other plate.

1. Initially, the plate with a charge of Q will have a uniform distribution of positive charge on its surface, while the plate with a charge of -q will have a uniform distribution of negative charge on its surface. This is because charges on a conductor will always distribute themselves evenly due to the repulsion of like charges.

Once equilibrium is reached, the electric field between the plates will be zero, as you correctly stated. This is because the electric field lines will cancel each other out due to the opposite charges on the plates. On the outer sides of the plates, the electric field will be non-zero and will point away from the positively charged plate and towards the negatively charged plate. This is because the electric field lines will originate from the positive charges and terminate on the negative charges.

2. When you inject an extra amount of charge m to the plate with an initial charge of -q, the charge distribution on that plate will change. This is because the additional charge will cause a redistribution of the charges on the surface of the plate. The negative charge will now be distributed more evenly across the surface, with a higher concentration near the injected charge. This is because the negative charges will repel each other and spread out to minimize their potential energy.

The electric field between the plates will also change, as the additional charge will create an imbalance between the two plates. This will result in a non-zero electric field between the plates, with a direction from the plate with the higher negative charge to the plate with the lower negative charge. On the outer sides of the plates, the electric field will still point away from the positively charged plate and towards the negatively charged plate.

To find the unique charge distribution, you can use the method of charge conservation. This means that the total charge on each plate before and after the injection of charge m must be the same. You can also use the method of potential energy minimization, where the distribution of charges will be such that the potential energy of the system is at a minimum. As for the polarization of the plates due to the other plate, this will depend on the dielectric properties of the material of the plates. If the plates are made of a non-conducting material, they may become polarized due to the presence of the opposite charges on the other plate. This will result in an additional electric field between the plates.

## 1. How does the distribution of electric charges on two parallel plates affect the electric field between them?

The distribution of electric charges on the two parallel plates determines the strength and direction of the electric field between them. If the charges are evenly distributed on the plates, the electric field will be uniform and perpendicular to the plates. If the charges are unevenly distributed, the electric field will be stronger in areas with higher charge density.

## 2. Can the distribution of electric charges on two parallel plates be manipulated?

Yes, the distribution of electric charges on two parallel plates can be manipulated by applying an external electric field. This can be done by connecting the plates to a power source, such as a battery, which will redistribute the charges on the plates.

## 3. How does the distance between the two parallel plates affect the distribution of electric charges?

The distance between the two parallel plates does not directly affect the distribution of electric charges. However, it does affect the strength of the electric field between the plates. The closer the plates are, the stronger the electric field will be, and vice versa.

## 4. What factors determine the amount of charge on each plate in a parallel plate capacitor?

The amount of charge on each plate in a parallel plate capacitor is determined by the voltage applied to the plates and the capacitance of the capacitor. The capacitance is determined by the distance between the plates, the area of the plates, and the material between the plates.

## 5. How does the shape of the parallel plates affect the distribution of electric charges?

The shape of the parallel plates does not directly affect the distribution of electric charges. However, it can indirectly affect the distribution by affecting the capacitance of the capacitor. For example, if the plates are curved, the distance between them may vary, which will affect the capacitance and therefore the distribution of charges.

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