Electric Field and Uniformly Charged Planes (conceptual)

[dimpledur]

Homework Statement

I am pursuing conceptual help regarding electric field due to uniformly charged planes.

Homework Equations

[tex]E=\frac{\sigma}{2\epsilon} [/tex]

The Attempt at a Solution

I understand that a capictor has plates that are +/-Q. However, how would you calculate the elctric field between the plates if the plates are not equal in charge? Would the electric field between the plates still be uniform?

[tex]E=\frac{\sigma _1}{2\epsilon} + \frac{\sigma _2}{2\epsilon} [/tex]

 

[Mandeep Deka]

There's a simple way to find that out!
Consider an arbitrary charge distribution on the two plates, of the capacitor. Find out the expression for the net electric field, at any point on the interior of a plate. What should it be equal to?

 

[dimpledur]

The force on a test charge q, due to plate Q1, would be:

F= (kqQ1)/r^2

F=[kq(sigma1)A]/r^2

therefore, E=[kq(sigma)A]/(r^2*q) =[k(sigma1)A]/r^2

If there was an addition plate with charge Q2, we would have E=[k(sigma1)A]/r^2

Therefore, the electric field due to the two plates would be:
E=[k(sigma1)A]/r^2 + [k(sigma1)A]/r^2

 

[dimpledur]

Does that look correct?

 

[rwisz]

= \frac{Q_1}{2A\epsilon} + \frac{Q_2}{2A\epsilon} = \frac{Q_1+Q_2}{2A\epsilon}

In order to calculate the electric field between two plates that are not equal in charge, you would use the equation E=\frac{\sigma}{2\epsilon}, where \sigma is the surface charge density and \epsilon is the permittivity of the medium. The electric field between the plates will still be uniform, as long as the plates are parallel and the distance between them is small compared to their size.

However, if the plates have different charges, the electric field between them will not be the same. The electric field will be stronger near the plate with the higher charge and weaker near the plate with the lower charge. This is because the electric field is directly proportional to the charge, so a higher charge will result in a stronger electric field.

To calculate the total electric field between the plates, you can add the individual electric fields from each plate. In the equation, \frac{\sigma _1}{2\epsilon} represents the electric field from the first plate and \frac{\sigma _2}{2\epsilon} represents the electric field from the second plate. So, the total electric field would be the sum of these two terms, which is \frac{Q_1+Q_2}{2A\epsilon}.

It is important to note that this equation assumes that the plates are infinite in size. If the plates are not infinite, then the electric field will not be completely uniform and will vary slightly near the edges of the plates. In this case, a more complex equation would be needed to accurately calculate the electric field between the plates.