# Homework Help: Calculate electric field at position x given ρ and thickness d?

1. Feb 19, 2013

### asdf12312

Calculate electric field at position x given ρ and thickness d??

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

Two large uniformly charged slabs, I and II, are stacked together as shown in the picture. Suppose the thickness of each slab is d= 0.2 μm. The volume charge densities in the negatively charged slab (I) and the positively charged slab (II) are −ρ and +ρ, respectively, with ρ = 8×10−3 C/cm3. The origin of the x-axis (i.e., x = 0) is at the interface of the two slabs. Suppose the slabs extend to infinity in the y and z-directions.

Using Gauss' law, the electric field outside of the two slabs region is: 0 N/C (this answer i got rite: E=pD/2e0 - pD/2e0= 0N/C

Apply Gauss' law to slab I to find the strength of the electric field at point A at position x=-0.199 μm?
Apply Gauss' law to slab II to find the strength of the electric field at point B at position x=0.101 μm?

2. Relevant equations
E (outside) = += pD/2e0 where e0=9x10^-12

3. The attempt at a solution

E(inside) at point A = pd/e0-(-pd/2e0)= 3pd/2e0
still not sure how to find d (it's different from D, i think.)

plaese help me with this. i hav a quiz on this tomorrow >.>

2. Feb 19, 2013

### Staff: Mentor

Re: Calculate electric field at position x given ρ and thickness d??

Not quite sure what you're doing here. Just apply Gauss's law. What would be your Gaussian surface? (You've correctly found that the field outside the slabs is zero.)

3. Feb 19, 2013

### asdf12312

Re: Calculate electric field at position x given ρ and thickness d??

gauss's law stats flux of a field is integral of EdA, or Q(in)/e0, which is what my teacher said.

in calculating the electric field inside the surface, according to my notes flux=EA+E(in)A, and enclosed charges are Q(in)=pV'=pAd

i equated gaussian surface eq. with this one:
Q(in)/e0=EA+E(in)*A
E(in)=(pd/e0)-E(out)

found that E(out) of the left slab is equal to pD/2e0 so:
E(in)=pd/e0-(pd/2e0)
E(in)=3pd/2e0

this is how i derived the equation but i hav a feeling idid it all wrong :(

4. Feb 19, 2013

### Staff: Mentor

Re: Calculate electric field at position x given ρ and thickness d??

OK, assuming one side is outside the slab and the other is at the point of interest.

So what does EA equal? I assume this is for the surface outside the slab.
So far, so good.

You already determined that the field outside is zero. Use that fact.

5. Feb 19, 2013

### asdf12312

Re: Calculate electric field at position x given ρ and thickness d??

well i mean the field outside both slabs is zero, but is that the same as the field outside the left slab? so that leaves pd/e0.

alright, but how do i find d now? i am guessing it's not the same as in the problem but even if it is:
(8*10^-3)(2*10^-7)/(9*10^-12)=177.77 N/C

6. Feb 19, 2013

### Staff: Mentor

Re: Calculate electric field at position x given ρ and thickness d??

Sure.
Right.
The thickness of your Gaussian region is the distance between the edge of the slab and the point A. You are given all that you need to calculate that distance.

7. Feb 19, 2013

### asdf12312

Re: Calculate electric field at position x given ρ and thickness d??

ok, that's what i thought.

d-x= 0.2 μm-0.199 μm = 1x10^-9 m

this is rite?

8. Feb 19, 2013

### Staff: Mentor

Re: Calculate electric field at position x given ρ and thickness d??

Yes, that is right.

9. Feb 19, 2013

### asdf12312

Re: Calculate electric field at position x given ρ and thickness d??

ok can you help me with 1 last thing? the answer is given as 9.05×10^5 N/C for x=-0.199 μm

i did (8*10^-3 C/cm3)(1*10^-9m)/(9x10^-12) and got 0.88 N/C. and i figured out to get the correct answer i need to multiply my answer by 10^6 (this is same for part 2 also), any idea why this is?

Last edited: Feb 19, 2013
10. Feb 19, 2013

### Staff: Mentor

Re: Calculate electric field at position x given ρ and thickness d??

You need to convert the charge density to standard units, C/m3. That will give you a factor of 10^6.