Outside of two oppositely charged infinite planes

[athenad07]

Can someone prove that outside of two infinite planes with opposite charge, the E-field got cancelled? But the between are not.

 

[Baluncore]

It does not get cancelled, it is simply not there.
Assuming the planes are conductors, the planes become equipotentials.
Between the planes, the boundary conditions have opposite charge, so there is an electric field gradient between them.
Outside the planes, the boundary conditions are a constant potential, so there can be no electric field gradient.

 

[Orodruin]

It does not get cancelled, it is simply not there.
Assuming the planes are conductors, the planes become equipotentials.
Between the planes, the boundary conditions have opposite charge, so there is an electric field gradient between them.
Outside the planes, the boundary conditions are a constant potential, so there can be no electric field gradient.

This is not correct. The setup allows a constant electric field orthogonal to the plates outside even if the planes are conducting. To avoid this, it is necessary to specify boundary conditions at infinity - ie the field vanishing.

The OP also does not mention anything about the planes being conductors. The planes will being equipotential boils down to symmetry.

I also do think it is correct to say that the field contributions from either plate cancel. A field, such as the EM field, by definition takes a value everywhere and therefore exists everywhere. Even if that value is zero, the field exists and is there.

 

[Baluncore]

The setup allows a constant electric field orthogonal to the plates outside even if the planes are conducting.

Plates or planes ?

 

[SredniVashtar]

The way I see it , this is an unphysical situation due to the fact that infinite planes are... impossible. But when used as a simplification device...

Consider one charged infinite plane: the field is constant in magnitude and direction (orthogonal) on both sides of the plane. If positively charged, the field emanates from the plane; if negatively charged it will impinge on the plane.

When you put two such planes near each other, the field between them will add up, while outside them will cancel out.

It's plain and simple principle of superposition.

 

[Orodruin]

Plates or planes ?

The OP says planes, which I realized half way through writing. Must have missed one occurrence. Regardless, that is hardly the relevant point.

The way I see it , this is an unphysical situation due to the fact that infinite planes are... impossible. But when used as a simplification device...

This is very often the case. Infinite planes, infinite wires, spherical cows, they are clearly idealised setups so I am not sure this needs to be explicitly pointed out. In particular when the setup is provided.

Consider one charged infinite plane: the field is constant in magnitude and direction (orthogonal) on both sides of the plane. If positively charged, the field emanates from the plane; if negatively charged it will impinge on the plane.

When you put two such planes near each other, the field between them will add up, while outside them will cancel out.

It's plain and simple principle of superposition.

… assuming homogenous boundary conditions at infinity.

 

[athenad07]

All the information that I have is these are two opposite sign charge infinite plane with some distance d. I also throw this question to my professor, he suggests me to prove it with Taylor series.

But I does not know what variable to setup this with, if anyone have any idea, can you please write down the full process that I can look it through? Thx

 

[SredniVashtar]

Use symmetry to establish the direction of the electric field.
Then use Gauss' theorem to find the magnitude.

 

[Charles Link]

It may worthwhile to consider $\nabla \times E=0$ for the electric field, so that for plates that are large but not completely infinite, you can do a complete path integral around and through the plates. You then get $\oint E \cdot dl=0$ over the closed loop by Stokes theorem, so that if you get $V=Ed$ between the plates, you will get exactly this same voltage outside the plates. Thereby $E$ may be nearly zero outside the plates, but not exactly zero. I believe my logic is correct here.

To shorten things, I used $E$ for the vector $\vec{E}$.

To add a little detail, we take the integral $\int \nabla \times E \cdot dA$ over an area in the plane of a loop we draw around and through the plates, and then apply Stokes theorem.