Evaluate the electric field for two slabs of charge

[zelscore]

Homework Statement

Two slab charges of thickness a with volume charge densities ±ρv are shown in
Fig. 2.30. Assume that the slabs are infinitely extended in the y- and z-directions.
Evaluate the electric field versus x.

Relevant Equations

Fig 2.30: https://imgur.com/a/TOgWvML
Hint: There is no field variation in the y- and z-directions. The electric field has
an x-component. Use Gauss’s law.

Thus I assume that one slab has positive charge Q1
and the other slab has negative charge Q2 = -Q1

There are 4 cases for the electric field:
1. x <= -a
2. -a <= x <= 0
3. 0 <= x <= a
4. a <= x

The general case:

Charge Density $\rho = \frac {Q} {V}$

Flux of E $\phi_e = \oint \vec E \cdot d \vec A = \frac {Q} {e_0} = EA ---> E = \frac {\rho} {e_0}*h$ where h is the length of the gaussian cylinder used in my case.

For case 2 and 3, h is (x+a) and (-x+a) respectively (thus x ranges from 0 to top or 0 to bottom)For case 1 and 4, the E field must be 0 because there is no charges there (charge density = 0)

This answer is in line with what the textbook says, however, if you test for x = 0 for both cases 2 and 3, you get that h = a (h = a is the highest h can be, thus E is too) which leads me to believe that the E field is the strongest inbetween the slabs, but this does not make sense to me because, spontaneously it feels like the E-field should be the strongest in the middle of the negatively(!) charged slab (reasoning: the +field will try to push a positive charge q placed inbetween the slabs in negative x-direction, and the -field will pull it in negative x-direction too, so the middle of the negative slab seems nice because then the +field has pushed as much as it could in the negative x-direction, while both ends of the -field will pull with equal force, cancelling each other out, so the charge will stay put)Edit: I think I understood it after some more thinking: If you put that charge in the middle of the slabs, the positive field will want to push a lot, while the negative field will want to pull a lot. So it makes sense that the E field is the strongest in the middle.SVARA

 

[BvU]

 

[rude man]

For case 1 and 4, the E field must be 0 because there is no charges there (charge density = 0)

This is true here beause of symmetry; just be careful not to assume that an E field is zero just because the enclosed charge is zero. The same E field can enter and leave the gaussian surfaces!

Good work!