Laplace's Equation and the potential above the xy-plane

[SquidgyGuff]

Homework Statement

Essentially it gives the potential above the xy-plane as

and I am tasked with verifying it satisfies laplace's equation, determining the electric field, and describing the charge distribution on the plane.

Homework Equations

then

The Attempt at a Solution

As far as I can tell it does not satisfy laplace's equation because when I take the gradient of the potential twice I get

which is non-zero. So taking the second equation up there I found

and using the third I found

.

In the case of the electric field I would assume the field in the x direction wouldn't matter (or at least shouldn't be there) because it is parallel to the field, but I'm not really sure if I'm doing this correctly.

 

[SteamKing]

Homework Statement

Essentially it gives the potential above the xy-plane as

and I am tasked with verifying it satisfies laplace's equation, determining the electric field, and describing the charge distribution on the plane.

Homework Equations

then

The Attempt at a Solution

As far as I can tell it does not satisfy laplace's equation because when I take the gradient of the potential twice I get

which is non-zero. So taking the second equation up there I found

and using the third I found

.

In the case of the electric field I would assume the field in the x direction wouldn't matter (or at least shouldn't be there) because it is parallel to the field, but I'm not really sure if I'm doing this correctly.

I think you are missing something here.

The Laplacian operator does not return a vector; it returns a scalar. The function φ here is a scalar function.

https://en.wikipedia.org/wiki/Laplace's_equation

 

[SquidgyGuff]

I think you are missing something here.

The Laplacian operator does not return a vector; it returns a scalar. The function φ here is a scalar function.

https://en.wikipedia.org/wiki/Laplace's_equation

But when you take the gradient isn't it this?

So taking the couble gradient would be:

Or is the laplacian not strictly the double gradient? If there's no directionality then it does equal zero and there for the charge distribution on the plane is 0 and the field is 0 right?

 

[SteamKing]

But when you take the gradient isn't it this?

So taking the couble gradient would be:

Or is the laplacian not strictly the double gradient? If there's no directionality then it does equal zero and there for the charge distribution on the plane is 0 and the field is 0 right?

The Laplacian is defined as the divergence of the gradient, or Δφ = div grad φ = ∇ ⋅ ∇φ = ∇²φ

 

[SquidgyGuff]

The Laplacian is defined as the divergence of the gradient, or Δφ = div grad φ = ∇ ⋅ ∇φ = ∇²φ

 

[SquidgyGuff]

The Laplacian is defined as the divergence of the gradient, or Δφ = div grad φ = ∇ ⋅ ∇φ = ∇²φ

Sorry about that last reply. Okay I get it now, so

which is then equal to

. So would that mean the the charge distribution is 0?

 

[SteamKing]

Sorry about that last reply. Okay I get it now, so

which is then equal to

. So would that mean the the charge distribution is 0?

It would seem so.

I know a little about the math behind potential theory, but I'm not au fait with electrostatics. If anyone else wants to jump in here ...

 

[Yosty22]

Would that mean the charge distribution is zero?

Yes.

A good way to check this is to apply Gauss' law to the electric field you calculated. It is extremely simple to show that the divergence of E goes to zero in this case, so that automatically implies that net charge is zero.

You stated in your initial post that the field above the xy-plane is given. Gauss' law allows us to create a Gaussian surface over the area of this region, so showing that ∇⋅E goes to zero shows us that there is no charge enclosed in the surface. This simply implies that the electric field (and thus the potential) is due to some charge distribution outside of the specified region or is simply an external field that is applied.

Gauss' law can be tricky like this sometimes, but it is helpful to think of the integral form, noting that ∫E⋅dA = Q_Enclosed/ε₀. This integral is around a surface (the Gaussian surface) so when you use the law to calculate the total charge, you are finding the total charge enclosed in the region. So what you have shown (correctly) is that the total charge distribution in this region is zero.

Note: If any of this isn't explained well, others feel free to jump in and correct me or elaborate.[/QUOTE]

 

[SquidgyGuff]

Would that mean the charge distribution is zero?

Yes.

A good way to check this is to apply Gauss' law to the electric field you calculated. It is extremely simple to show that the divergence of E goes to zero in this case, so that automatically implies that net charge is zero.

You stated in your initial post that the field above the xy-plane is given. Gauss' law allows us to create a Gaussian surface over the area of this region, so showing that ∇⋅E goes to zero shows us that there is no charge enclosed in the surface. This simply implies that the electric field (and thus the potential) is due to some charge distribution outside of the specified region or is simply an external field that is applied.

Gauss' law can be tricky like this sometimes, but it is helpful to think of the integral form, noting that ∫E⋅dA = Q_Enclosed/ε₀. This integral is around a surface (the Gaussian surface) so when you use the law to calculate the total charge, you are finding the total charge enclosed in the region. So what you have shown (correctly) is that the total charge distribution in this region is zero.

Note: If any of this isn't explained well, others feel free to jump in and correct me or elaborate.

[/QUOTE]
Thank y'all so much! Just needed to make sure my reasoning was sound :)