# Electric field of two plates separated by a spacial charge.

1. Jun 18, 2017

### Decimal

Hello,

I think I solved the question, but I am confused by the elaboration on the exercise given by my professor.

1. The problem statement, all variables and given/known data

A very large plate charged with charge Q is placed between x = -d and x=0. A similar plate charged with charge 3*Q/2 is placed between x = 3*s and x = 3*s + d. Here, s and d are given constants.

Between the plates exists a spacial charge ρ, between x = 0 and x = 3s. Consider two points P and S placed on either side of the two plates on the x-axis, but also very close to them. Point P is placed at Xp > 3*s+d, and Point S at Xs < -d.

Calculate the electric field in point P and S. The plates are large enough that side effects can be ignored and the electric field will only have a component parallel to the x-axis.

2. Relevant equations

Gauss law for electric fields

3. The attempt at a solution

This is what I did:

Because these plates are very large, the electric field is not dependent on distance from the plates and the electric field will be the same size on either side, just in the opposite direction.

Take a gauss cylinder through the plates with the two sides through point S and P.

Ep⋅dA + ∫Es⋅dA = Qenc/ε0

Then since Ep and Es point in the same direction as dA in both cases you can write

∫Ep*dA + ∫Es*dA = Qenc/ε0

I will call the surface of the cylinders side A'

Ep*A' + Es*A'=Qenc/ε0

Now this is where my problem arises. I know how to calculate Qenc and work out this equation, but in the elaborations given by my professor the equation becomes:

Ep*A' - Es*A'=Qenc/ε0

I don't understand where this minus comes from. dA and Es should point in the same direction right?

2. Jun 18, 2017

### Staff: Mentor

It depends on your definition of the signs. If you want both electric fields to follow the same sign convention (probably: positive=to the right), then the areas have to be A' and -A' because their orientation is different.