Tangential electric force at a surface

[josephsanders]

Suppose you have an infinite plane of charge. If the surface charge density is uniform, would the tangential electric force always be zero, even if it is not a conductor nor static? My thought process for this is that if you look at each point charge and draw the electric field lines, then at any point on the surface the electric field in the tangential direction should cancel by symmetry?

I also have a followup question. I know by Maxwell's equations, the tangential electric force is continuous for any surface. However, this doesn't make much sense to me because the tangential fields on each side of the surface are by definition parallel to each other. So does it even make sense to say they are continuous or not?

Thank you for taking the time to read my post!

 

[TSny]

Suppose you have an infinite plane of charge. If the surface charge density is uniform, would the tangential electric force always be zero, even if it is not a conductor nor static? My thought process for this is that if you look at each point charge and draw the electric field lines, then at any point on the surface the electric field in the tangential direction should cancel by symmetry?

It seems reasonable to take the tangential component of E to be zero so that the direction of E at any point off of the plane is perpendicular to the plane.

However, you could add to this field a uniform electric field parallel to the infinite plane and the total E field would still satisfy Maxwell's equations! Seems strange, but Maxwell's equations are differential equations and a particular solution to the equations depends on the specification of boundary conditions for the field.

I also have a followup question. I know by Maxwell's equations, the tangential electric force is continuous for any surface. However, this doesn't make much sense to me because the tangential fields on each side of the surface are by definition parallel to each other. So does it even make sense to say they are continuous or not?

Let the z-axis be perpendicular to the plane so that the x and y axes are parallel to the plane. It does not follow from the definition of "tangential field" that the tangential fields on each side of the plane must be parallel to each other. If the tangential component of the field on one side is in the x-direction while the tangential field on the other side is in the y-direction, then you would have tangential fields that are not parallel to each other.

Even if the tangential fields on each side are parallel, they might not have the same magnitude. Then, the tangential field would not be continuous across the surface.

 

[vanhees71]

Your charge distribution can be described by
$$\rho(\vec{x})=\sigma \delta(x_3),$$
where $\sigma=\text{const}$ is your uniform surface-charge density.

By symmetry the electrostatic potential should only be a function of $x_3$. Then the Poisson equation reads
$$\Delta \Phi =\Phi''(x_3) = -\frac{\sigma}{\epsilon_0} \delta(x_3).$$
For $x_3 \neq 0$ solution obviously is
$$\Phi(x_3)=A x_3 +B,$$
where $A$ and $B$ take different values for $x_3<0$ and $x_3>0$ to get the singularities to make up the $\delta$ distribution on the right-hand side of the equation. We can set $B=0$ on both sides, because this still satisfies the continuity condition for $\Phi$ at $x_3=0$. So we get
$$\Phi(x_3)=A_< x_3 \Theta(-x_3) + A_> x_3 \Theta(x_3).$$
Integrating the Poisson equation over a little interval $(-\epsilon,\epsilon)$ and then letting $\epsilon \rightarrow 0^+$ leads to
$$\Phi'(0^+)-\Phi'^(-0^+)=A_>-A_<=-\frac{\sigma}{\epsilon_0}.$$
By reflection symmetry $\vec{E}(x_3)=-\vec{E}(-\vec{x}_3)$, which means $A_<=-A_>$, you get
$$A_>=-A_<=-\frac{\sigma}{2 \epsilon_0}$$
and thus
$$\vec{E}=\vec{e}_3 \frac{\sigma}{2 \epsilon_0} [\Theta(x_3)-\Theta(-x_3)].$$

 

[josephsanders]

It seems reasonable to take the tangential component of E to be zero so that the direction of E at any point off of the plane is perpendicular to the plane.

However, you could add to this field a uniform electric field parallel to the infinite plane and the total E field would still satisfy Maxwell's equations! Seems strange, but Maxwell's equations are differential equations and a particular solution to the equations depends on the specification of boundary conditions for the field.Let the z-axis be perpendicular to the plane so that the x and y axes are parallel to the plane. It does not follow from the definition of "tangential field" that the tangential fields on each side of the plane must be parallel to each other. If the tangential component of the field on one side is in the x-direction while the tangential field on the other side is in the y-direction, then you would have tangential fields that are not parallel to each other.

Even if the tangential fields on each side are parallel, they might not have the same magnitude. Then, the tangential field would not be continuous across the surface.

Thank you so much! This really helped clear my head about it :)