How Does Gauss' Law Apply to an Infinite Charged Wall?

[RobTwox]

Homework Statement

[/B]
Assume that the infinite wall −2 < y < 2 is filled with a uniform charge of density ρ_v = 3 C/m3. Obtain an expression for Ey at a general point (0,y,0) and plot Ey as a function of y over −5 < y < 5.

Homework Equations

E = ρ_v / ( 4 π ε₀ ρ)

∫∫∫D⋅ds = Qenc

The Attempt at a Solution

I will be honest. I am not really sure where to begin. I am not concerned about the plot. I need to know how to set this up. I am thinking gauss law or columns law. If I use coloumns law I set up a triple integral that will go to nfinity due to the limits of x and z. Gauss law seems like the way to go. If this is the case, do I wrap my gausian surface around a chunk of the wall, or do I take some segment? I need lost of help on this. I honestly do not know where to begin

 

[RobTwox]

Please, someone help me out here. If only with the set up

 

[TSny]

Gauss' law is the way to go.

Before trying to set up any equations, use the symmetry of the problem to determine the direction of E for any point inside or outside the slab (wall).

Also, determine all points where E = 0.

 

[RobTwox]

Gauss' law is the way to go.

Before trying to set up any equations, use the symmetry of the problem to determine the direction of E for any point inside or outside the slab (wall).

Also, determine all points where E = 0.

because the charge is positive it points outward at all points. the only place I think it might be 0 is at the center. I think that here it is equal and opposite and therefor 0

 

[RobTwox]

because the charge is positive it points outward at all points. the only place I think it might be 0 is at the center. I think that here it is equal and opposite and therefor 0

Is this correct?

 

[RobTwox]

Can anybody help me on this, I am running out of time fast

 

[J Hann]

Note that outside of the wall (since the electric field lines cannot diverge because the wall is infinite) the field
must be constant everywhere outside of the wall. This should provide some hints at what is happening inside
of the wall.

 

[RobTwox]

Note that outside of the wall (since the electric field lines cannot diverge because the wall is infinite) the field
must be constant everywhere outside of the wall. This should provide some hints at what is happening inside
of the wall.

inside I suspect it is 0 at the middle. as I move outward in either direction it would begin to rise.

My main concern is getting the expression. From that I can plot it.

Can you help me. where do I start?

 

[TSny]

inside I suspect it is 0 at the middle. as I move outward in either direction it would begin to rise.

My main concern is getting the expression. From that I can plot it.

Can you help me. where do I start?

Yes, the field is zero everywhere on the plane y = 0.
Try a Gaussian surface in the shape of a rectangular box with one face at y = 0.

 

[RobTwox]

Yes, the field is zero everywhere on the plane y = 0.
Try a Gaussian surface in the shape of a rectangular box with one face at y = 0.

so my Qenclosed = ρ_v L H where L and H are edges of the box
and ∫∫∫D⋅dv = DLH
can I say D= ρ_v
and then
E = ρ_v / ε0

 

[RobTwox]

I'm really lost with this. We are just covering gauss law this is my first problem

 

[TSny]

so my Qenclosed = ρ_v L H where L and H are edges of the box

ρ is a volume charge density. So, to get the charge enclosed, you need to multiply ρ by a volume. Is L H a volume?

and ∫∫∫D⋅dv = DLH

In Gauss' law, the integral of the field is a surface integral representing flux through the surface. Which side or sides of the box will have nonzero flux?

can I say D= ρ_v
and then
E = ρ_v / ε0

No.

 

[RobTwox]

ρ is a volume charge density. So, to get the charge enclosed, you need to multiply ρ by a volume. Is L H a volume?

In Gauss' law, the integral of the field is a surface integral representing flux through the surface. Which side or sides of the box will have nonzero flux?No.

ok if 1 face is at y =0 then that face has no flux through. The only face we care about is the one poking out the side (that's where we are asked to find E) but the top and bottom would have flux too.
so Qenc =ρ_v L W H
and then I'm not sure what to do with the integral if I do ∫d⋅dv then I get D L W H and I get the same expression E = ρv / ε0
I don't think this is right

 

[TSny]

I'm not sure what direction your y-axis is pointing. I'm guessing that your y-axis is running horizontally so that y = 0 is a vertical plane in the middle of the vertical wall of charge. So, we have a Gaussian box placed such that one face of the box is at y = 0. As you say, there will not be any flux through this face because the field is zero on the this face. There are 5 other faces. Which of these faces has nonzero flux?

Your expression Qenc =ρv L W H will be correct if you interpret L W and H properly. You will need to deal separately with 2 cases: (1) finding the field at a point inside the wall, and (2) finding the field at a point outside the wall.

 

[RobTwox]

I'm not sure what direction your y-axis is pointing. I'm guessing that your y-axis is running horizontally so that y = 0 is a vertical plane in the middle of the vertical wall of charge. So, we have a Gaussian box placed such that one face of the box is at y = 0. As you say, there will not be any flux through this face because the field is zero on the this face. There are 5 other faces. Which of these faces has nonzero flux?

Your expression Qenc =ρv L W H will be correct if you interpret L W and H properly. You will need to deal separately with 2 cases: (1) finding the field at a point inside the wall, and (2) finding the field at a point outside the wall.

lets deal with case 1 first where the box is completely inside. at this point the outside face will be the only one with flux thru. the flux will be parallel to all other faces. Am I right?

 

[TSny]

lets deal with case 1 first where the box is completely inside. at this point the outside face will be the only one with flux thru. the flux will be parallel to all other faces. Am I right?

Yes. Good.

 

[RobTwox]

Yes. Good.

ok then I need to set up ∫D⋅ds
isn't this just D * surface area or D L H

 

[TSny]

Yes.

 

[RobTwox]

Yes.

alrighty then so if I set eqns equal to each other D= ρ_v W / ε0

wont the expression be the same for the outside of wall case?

I know I have to get rid of the w this is not given how do I do this?

 

[TSny]

How is W related to the value of y for the face which had nonzero flux?

For points outside the wall, your box will only be partially filled with charge.

 

[RobTwox]

How is W related to the value of y for the face which had nonzero flux?

For points outside the wall, your box will only be partially filled with charge.

w = 2
so
E= 2 ρv / ε0

I don't understand how a partially filled box will change things. Dosent the flux still come through just the same?

 

[TSny]

w = 2

When finding the field inside the wall, you should choose one of the faces of the box to be at an arbitrary distance y < 2 from the y = 0 plane. So, W for this box would not be W = 2. Express W in terms of y.

I don't understand how a partially filled box will change things. doesn't the flux still come through just the same?

For finding the field outside the wall, you will want to construct your box so that one face is at some arbitrary distance y > 2.
How are you going to find the total charge enclosed in this box?

 

[RobTwox]

When finding the field inside the wall, you should choose one of the faces of the box to be at an arbitrary distance y < 2 from the y = 0 plane. So, W for this box would not be W = 2. Express W in terms of y.

would say -2 < y < 2 ? I'm not sur how to express in terms of y
For finding the field outside the wall, you will want to construct your box so that one face is at some arbitrary distance y > 2.
How are you going to find the total charge enclosed in this box?

ρ_v * Volume = ρ_v L W H

but I guess if the box is half full I could say

ρ_v L W H / 2Is any of this right?

 

[TSny]

Well, I think we're getting lost in jumping around. I think you need to finish the result for inside the wall before going to outside.

 

[RobTwox]

Well, I think we're getting lost in jumping around. I think you need to finish the result for inside the wall before going to outside.

ok, so if I express w in terms of y is it
E= y ρv / ε0

 

[TSny]

OK, good.

So, going to points outside the wall. Now let y be some value greater than 2. So your box has a face at y = 0 and a parallel face at some value of y outside the wall. How are you going to express the total charge inside the box?

 

[RobTwox]

ok, so if I express w in terms of y is it
E= y ρv / ε0

OK, good.

So, going to points outside the wall. Now let y be some value greater than 2. So your box has a face at y = 0 and a parallel face at some value of y outside the wall. How are you going to express the total charge inside the box?

Qenc = ρ_v * volume.
volume is y * L * H and y =2
so Qenc = ρ_v 2 H L

 

[TSny]

Right.

How do you express the total flux through the surface of the box?

 

[RobTwox]

Right.

How do you express the total flux through the surface of the box?

this is D⋅ds so

D L H

 

[TSny]

Yes. Note that the D here corresponds to the field at an arbitrary point where y is greater than 2. Putting it together, what do you find for D (or E) at an arbitrary point outside the wall?