Electrostatics - parallel plate capacitor

[apjack]

Two parallel uniform charged circular conducting metal disks contains a charge of +Q and -Q on the inner surfaces, and little amounts of charge +q and -q on the outer surfaces. The detail is shown in the diagram.
Each plate has a radius R and thickness t. The separation of the plates is d.

Question: How much charge q is on the outside surface of the disk. The answer is in terms of Q and without the approach by Gauss's law.
Hint: E-field of a uniformly charged disk of radius R a distance z from the center of the disk is as following equation.

Please help to solve this problem. It is strange to me that the outer surface carries certain amount of charges.

 

[jbsteele]

The answer is q = (Q/2πRt) * (d - t) where Q is the charge on the inner surfaces, R is the radius of the plates, t is the thickness of the plates and d is the separation between the plates. This can be calculated by using the equation for the electric field of a uniformly charged disk, E = (Q/2πRz) * (1 - (z/R)) where Q is the charge on the disk, R is the radius of the disk and z is the distance from the center of the disk. Since the outer surface of each plate is a distance d - t from the center of the plate, we can calculate the electric field at this distance and use it to calculate the charge q.

 

[nlightner]

I would approach this problem by first understanding the concept of a parallel plate capacitor. A parallel plate capacitor consists of two parallel plates with opposite charges, separated by a distance. This creates an electric field between the plates, and the amount of charge on each plate determines the strength of the electric field.

In this scenario, we have two parallel charged disks, each with a radius R and thickness t. The inner surfaces of the disks have charges of +Q and -Q, respectively. The outer surfaces of the disks also have small amounts of charge, +q and -q. This is due to the fact that the electric field created by the inner charges extends beyond the disks and affects the outer surfaces.

To determine the amount of charge q on the outer surface of the disks, we can use the equation for the electric field of a uniformly charged disk. This equation states that the electric field at a distance z from the center of the disk is proportional to the charge density, which is the charge per unit area. This can be written as:

E = σ/ε0

Where E is the electric field, σ is the charge density, and ε0 is the permittivity of free space. We can rearrange this equation to solve for σ:

σ = ε0E

Now, we can use this equation to find the charge density on the outer surface of the disks. Since the electric field is created by the charges on the inner surfaces, we can use the known values of Q, R, and d to calculate the electric field between the plates. This can be done using the equation for the electric field of a parallel plate capacitor:

E = Q/ε0A

Where A is the area of the plates, which in this case is πR^2. Substituting this into the equation for charge density, we get:

σ = ε0(Q/ε0A) = Q/A

Now, we know that the outer surface of each disk has a charge density of Q/A. Since the area of the outer surface is also πR^2, we can use this to find the total charge on the outer surface:

q = (Q/A)(πR^2) = Q(πR^2)/A = Q(πR^2)/(πR^2) = Q

Therefore, the amount of charge on the outer surface of each disk is equal to the charge on the inner surface, Q. This makes