Finding E and V involving 2 planes.

[Mindscrape]

Keep in mind that

- \nabla V = E

Also, you have to keep in mind how integration bounds work, you go from the smallest to the largest. You don't even have to use bounds for this problem. You can simply put in the constant of integration, then once you decide on which plate you want to have zero potential, you can adjust the constants accordingly.

As it is, you're close, but you're integrations need some fine tuning. I know your struggling, but I don't want to hand you the answers, the struggle is good.

 

[xxbigelxx]

Hey Mindscrape.

Are you telling me that I am close, or noface, or both of us? I understand what you said, and I think that's what I tried to do.

 

[kuruman]

This is how you do it. I will do region I (upper region, z>6 cm) and you will have to do the others the same way. The net electric field in region I is E_I = 5/ε₀. To find the potential in region I, you start with the the known equation

V_I(z)-V(ref)=-\int^{z}_{ref}E_I dz

Here ref = 6 cm, i.e. V(6) = 0. The equation becomes

V_I(z)-0=-\int^{z}_{6}\frac{5}{\epsilon_0} dz

This gives you V_I(z). You can get the potential in the other two regions the same way.

 

[xxbigelxx]

Ok I think I got it now. I believe I followed everything you told me. How's it look?

 

[kuruman]

Looks fine.

 

[xxbigelxx]

Great. Do I leave them as distinct quantities, or can I combine them? The questions says "Determine V for all points" so I feel like there should be another form of the answer.

 

[kuruman]

The way to present the solution is by listing the potentials in each region, such as

V(x,y,z) = ... for x ≥ 6 cm (top region)
V(x,y,z) = ... for 6 cm ≥ x ≥ -4 cm (in-between region)
V(x,y,z) = ... for x ≤ -4 cm (bottom region)

 

[xxbigelxx]

Ok I see what you are saying. I think you meant to say z instead of x for all those bounds you gave, but I agree otherwise. And all of those units should be 'V' I believe as well.

Finally for the drawing of the equipotentials, they are just going to be planes parallel to the z planes right?

 

[noface]

Yeah, they're just planes with normal vector in the z direction. I'm confused about the middle region...why wouldn't the "-4" be involved at all in the bounds. I have the same thing for the other two regions and understand why, but I don't understand why the middle region wouldn't be limited by the physical boundary of -4.

 

[kuruman]

Ok I see what you are saying. I think you meant to say z instead of x for all those bounds you gave, but I agree otherwise. And all of those units should be 'V' I believe as well.

Sorry, yes I meant to say z instead of x for all the bounds.

 

[xxbigelxx]

I can't answer for sure, but I would guess it's just because you are only interested in the plate where the E is originating (in this case at z= 6cm)

 

[Mindscrape]

Yeah, they're just planes with normal vector in the z direction. I'm confused about the middle region...why wouldn't the "-4" be involved at all in the bounds. I have the same thing for the other two regions and understand why, but I don't understand why the middle region wouldn't be limited by the physical boundary of -4.

Because -4 isn't the reference point. If you wanted the capacitance then you would need the potential difference between V(6) and V(-4), otherwise it doesn't matter. Don't worry, the boundaries solve any issues of ambiguity.

Lastly, don't worry, this is a textbook problem. It's meant for you to learn about EM. In reality, with no approximations, you would always set the reference voltage at infinity. Griffiths has a good discussion of this.