Electric field of an infinite sheet of charge.

[SayWhat1]

Homework Statement

Using direct integration show that the electric field due to an infinite sheet with charge density σ is independent from the distance from the sheet and equals \frac{σ}{2\epsilon_{0}}

Homework Equations

\int\int k \frac{Q}{r^{3}}dxdy

Which should lead directly to the equation:

kQ\int\int\frac{z}{(x^{2}+y^{2}+z^{2})^{\frac{3}{2}}}dxdy

The Attempt at a Solution

We had to do the same thing with a line of constant charge and he told us we could use that result in this problem, which is basically the first half of the problem and gives you the result E=\frac{\lambda}{2\pi\epsilon_{0}z}

I don't know if my problem is with the surface integral or the physics but this is just baffling me. I can't get to the "middle" with what I start with and if I start with what he gave us, I can't get to the solution.

Edit: I got the answer the first time by getting rid of the double integral and thinking of it as an infinite number of concentric rings of charge and integrating that but he didn't want it that way. He wants it with the double integral.

 

[Muphrid]

kQ/r^3? Who told you that? It should be more like, say

\int \int k \frac{\sigma}{r^2} \; dx \; dy

 

[SayWhat1]

You need to account for the direction of the field so on the end of yours should be \frac{\hat{r}}{\left|\hat{r}\right|}. The magnitude of \hat{r} is gong to be r again, which gives you the r^{3} on the bottom.

 

[Chestermiller]

Homework Statement

Using direct integration show that the electric field due to an infinite sheet with charge density σ is independent from the distance from the sheet and equals \frac{σ}{2\epsilon_{0}}

Homework Equations

\int\int k \frac{Q}{r^{3}}dxdy

Which should lead directly to the equation:

kQ\int\int\frac{z}{(x^{2}+y^{2}+z^{2})^{\frac{3}{2}}}dxdy

The Attempt at a Solution

We had to do the same thing with a line of constant charge and he told us we could use that result in this problem, which is basically the first half of the problem and gives you the result E=\frac{\lambda}{2\pi\epsilon_{0}z}

I don't know if my problem is with the surface integral or the physics but this is just baffling me. I can't get to the "middle" with what I start with and if I start with what he gave us, I can't get to the solution.

Edit: I got the answer the first time by getting rid of the double integral and thinking of it as an infinite number of concentric rings of charge and integrating that but he didn't want it that way. He wants it with the double integral.

It is possible to do it his way by using a sequence of rectangular straight line elements of width dy oriented in the x direction, and integrating over these elements.

 

[TSny]

Also, make sure that when you use E = λ/(2\pi\epsilon_oz) for an infinitely long thin strip of the plane that you realize z here is not the perpendicular distance from the plane to the field point. It's the perpendicular distance from the strip to the field point. It might be better to write it as r rather than z.

 

[SayWhat1]

It is possible to do it his way by using a sequence of rectangular straight line elements of width dy oriented in the x direction, and integrating over these elements.

If I did it this way it'd reduce it down to a single integral again, and he specifically wants the double.

Also, make sure that when you use E = λ/(2\pi\epsilon_oz) for an infinitely long thin strip of the plane that you realize z here is not the perpendicular distance from the plane to the field point. It's the perpendicular distance from the strip to the field point. It might be better to write it as r rather than z.

I think this is were I'm hanging up. So when I do that, am I going to take it times the unit vector again, or is that all taken care of in the beginning? Maybe because it seems so simple I'm just psyching myself out because all I need to get rid of from that mid-step is a pi.

 

[Muphrid]

I think you should consider converting from rectangular to cylindrical coordinates. It makes the problem much easier, as one of the two integrals is trivial.

 

[SayWhat1]

I think you should consider converting from rectangular to cylindrical coordinates. It makes the problem much easier, as one of the two integrals is trivial.

I'm not entirely sure how to plug the cylindrical coordinates into my equation.

 

[TSny]

If I did it this way it'd reduce it down to a single integral again, and he specifically wants the double.



I think this is were I'm hanging up. So when I do that, am I going to take it times the unit vector again, or is that all taken care of in the beginning? Maybe because it seems so simple I'm just psyching myself out because all I need to get rid of from that mid-step is a pi.

Yes, you will need to take into account that the field from one of the strips will point directly away from the strip. Since you know the total field from all the strips will be in the z direction, you will need to take the z-component of the field of each strip and integrate this z-component for all the strips. This will be your y-integral from -∞ to ∞. This is what Chestermiller was hinting at.

When you start with the double integral you can first fix y and integrate over x from -∞ to ∞. But that will clearly just produce the field of a thin strip of thickness dy. So, rather than actually doing the integral over x, you can just use the given result for an infinite line of charge. You need to convince yourself that λ = σdy for the thin strip.

 

[Muphrid]

I'm not entirely sure how to plug the cylindrical coordinates into my equation.

What are x, y, z in terms of cylindrical coordinates?

 

[SayWhat1]

Yes, you will need to take into account that the field from one of the strips will point directly away from the strip. Since you know the total field from all the strips will be in the z direction, you will need to take the z-component of the field of each strip and integrate this z-component for all the strips. This will be your y-integral from -∞ to ∞. This is what Chestermiller was hinting at.

When you start with the double integral you can first fix y and integrate over x from -∞ to ∞. But that will clearly just produce the field of a thin strip of thickness dy. So, rather than actually doing the integral over x, you can just use the given result for an infinite line of charge. You need to convince yourself that λ = σdy for the thin strip.

Thank you, the light bulb just clicked on. I've been slaving over this thing for a week and I just couldn't get it to work out.

Thanks a lot!

What are x, y, z in terms of cylindrical coordinates?

Well, in the cylindrical coordinates p=\sqrt{x^{2}+y^{2}} and \psi=arcsin(\frac{y}{p}) and z stays the same. I'm just not sure how to represent r in terms of p \psi and z.

 

[Muphrid]

I mean, you want to say x is some function of the cylindrical coordinates--e.g. x = \rho \cos \psi. Same process for y and z.

The point is your integral turns into this:

\frac{\sigma}{4\pi \epsilon_0} \int_0^{2\pi} \int_0^\infty \frac{z}{(\rho^2 + z^2)^{3/2}} \rho \; d\rho \; d\psi

And it should be fairly obvious by this point that you can attack this with a variable substitution (see that d(\rho^2) = 2 \rho \; d\rho). Though I see through your addendum to the first post that you already did it through concentric rings and this form should be familiar anyway.

 

[SayWhat1]

I mean, you want to say x is some function of the cylindrical coordinates--e.g. x = \rho \cos \psi. Same process for y and z.

The point is your integral turns into this:

\frac{\sigma}{4\pi \epsilon_0} \int_0^{2\pi} \int_0^\infty \frac{z}{(\rho^2 + z^2)^{3/2}} \rho \; d\rho \; d\psi

And it should be fairly obvious by this point that you can attack this with a variable substitution (see that d(\rho^2) = 2 \rho \; d\rho). Though I see through your addendum to the first post that you already did it through concentric rings and this form should be familiar anyway.

That's kinda what I was afraid of when I started thinking about it like you were saying. I'll probably work it out like this too to see if I can get some extra credit or something. Thanks for all the help!