Electric field from a gaussian pillbox in an infinite slab of thickness 2d

[indigojoker]

The question states that there is an infinite plane slab of thickness 2d that has a uniform charge density rho. Find the electric field as a function of y, where y=0 at the center.

inside plane:
$$\int E da = 2EA = \frac{\rho A y}{\epsilon_o}$$

outside plane:
$$\int E da = 2EA = \frac{\rho A d}{\epsilon_o}$$

The 2d thickness runs along the y axis, and the plane is infinite along the x and y axis.

Is this the right setup?

 

[learningphysics]

everything looks right except I think you're missing a factor of 2... if y is measured from the center... then taking a section center from the center with width 2y... you'll have

$$2E_y*A = \frac{\rho A (2y)}{\epsilon_o}$$

same with outside the plane...

$$2E_y*A = \frac{\rho A (2d)}{\epsilon_o}$$

 

[indigojoker]

good point!

I don't know why the E-field would be negative then y < 0?

should the e-field always be positive when dealing with plates? Like for an infinite plane, the E field is:

$$E = \frac{\sigma}{2 \epsilon _o}$$

the E-field can't be negative (unless the charge on the plane was negative). So why does my expression give a negative E when i go to negative y?

 

[learningphysics]

The field at $$E_{-y} = -E_{y}$$ (where y>0). We took this into account when we did gauss law... that's why we got 2*Ey*A... It was Ey*A + (-Ey)*(-A)... (this is just integral of E.dA)

We were actually taking y to be positive and using the fact that $$E_{-y} = -E_y$$.

This makes sense by symmetry... if the field at y is upward... then the field at -y should be downward...

 

[learningphysics]

good point!

I don't know why the E-field would be negative then y < 0?

should the e-field always be positive when dealing with plates? Like for an infinite plane, the E field is:

No.

$$E = \frac{\sigma}{2 \epsilon _o}$$

the E-field can't be negative (unless the charge on the plane was negative). So why does my expression give a negative E when i go to negative y?

The field can be negative for that case... one side the field is upward... the other side the field is downward. You have to choose a particular direciton as positive... the other side is negative.

 

[indigojoker]

So in my case:

$$E = \frac{\rho d }{\epsilon_o} y$$ when y>d
and
$$E = \frac{-\rho d }{\epsilon_o} y$$ when y<d

where y is the unit vector in the y direction

 

[learningphysics]

So in my case:

$$E = \frac{\rho d }{\epsilon_o} y$$ when y>d
and
$$E = \frac{-\rho d }{\epsilon_o} y$$ when y<d

where y is the unit vector in the y direction

yeah. just to be sure though, you didn't use y as a unit vector here right: ?

$$2E_y*A = \frac{\rho A (2y)}{\epsilon_o}$$

here y is the distance from the center...

 

[indigojoker]

yes.

it seems like when I do the same thing for the opposite side, i get the same equation. is it the negative y unit vector that makes the E field negative?

 

[learningphysics]

yes.

it seems like when I do the same thing for the opposite side, i get the same equation. is it the negative y unit vector that makes the E field negative?

yes... By symmetry the field is in the opposite direction. Hence the vector needs to be negative...

 

[du.art]

I have a similar exercise and I'm finding some difficulty in understanding it.
So from Gauss' Law I have
$$\epsilon_{0}\oint EdA=q_{enc}$$
I know
$$q_{enc}=\rho\,V=\rho\,y$$
But I don't get where do those "2" come from as in 2EA and 2y.
I also fail to understand the difference between the inside and outside equations. The y gets replaced by d but the only connection I know of is that when y=2d it's the end of the plate