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

[asdf12312]

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

Homework Statement

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?

Homework Equations

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

E(inside)=pd/e0-E(outside) - not sure about this equation i derived

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 >.>

 

[Doc Al]

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.)

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.)

 

[asdf12312]

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
pAd/e0=EA+E(in)*A
E(in)*A=(pAd/e0)-EA
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 :(

 

[Doc Al]

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

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.

E(in)*A=(pAd/e0)-EA
E(in)=(pd/e0)-E(out)

So far, so good.

found that E(out) of the left slab is equal to pD/2e0 so:

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

 

[asdf12312]

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

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.

 

[Doc Al]

well i mean the field outside both slabs is zero, but is that the same as the field outside the left slab?

Sure.

so that leaves pd/e0.

Right.

alright, but how do i find d now?

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.

 

[asdf12312]

ok, that's what i thought.

d-x= 0.2 μm-0.199 μm = 1x10^-9 m

this is rite?

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.

 

[Doc Al]

ok, that's what i thought.

d-x= 0.2 μm-0.199 μm = 1x10^-9 m

this is rite?

Yes, that is right.

 

[asdf12312]

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?

 

[Doc Al]

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?

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