Electrostatic Boundary Conditions

[Hypercube]

Hello PF community,

I am currently self-studying electrodynamics from Griffiths textbook, and I'm at a point where the book discusses electrostatic boundary conditions. If someone can please check if my reasoning is right.

So, as I am approaching an infinite, uniformly charged plane (let the charge be positive), there is electric field pointing away from the plane, which has a magnitude equal to:

$$\vec E=\frac{\sigma}{2\epsilon_0}\hat n$$

where $\hat n$ is a vector perpendicular to the plane. This can be derived from Gauss' law, and it shows that electric field of a point does not depend on the distance from the plane. Magnitude on the other side of the plane would be the same, except it would have opposite direction.

However, according to the book, as I'm approaching the boundary and eventually cross it, there is a discontinuity. The author writes:

$$E_{above}-E_{below}=\frac{\sigma}{\epsilon_0}$$

Note that the author does not have these in bold. This implies that (somehow) the magnitude of the electric field above the plane is not equal to the magnitude of the electric field below it. How is that possible? I expected RHS to be 0 in this case.

Thanks.

 

[mfb]

Note that the author does not have these in bold.

Probably an oversight.

The magnitude can be different (and for an ideal plate capacitor the difference in magnitude will be this value, because the field on the outside is zero), but the boundary condition is the change in the field vector.

 

[jtbell]

Perhaps Griffiths was thinking in terms of a component of the field, e.g. $E_z$ for a sheet parallel to the xy-plane. Unlike the magnitude $E$, this can be positive or negative.

 

[Chandra Prayaga]

Perhaps Griffiths was thinking in terms of a component of the field, e.g. $E_z$ for a sheet parallel to the xy-plane. Unlike the magnitude $E$, this can be positive or negative.

Yes. Griffiths was in fact referring to the component of the field perpendicular to the surface. I happen to have a copy of his book handy and checked again, to make sure. It was neither an oversight, nor an imprecisely stated result. The relation given in Griffiths 3rd edition says:
E^⊥_above - E^⊥_below = σ/ε₀

 

[Hypercube]

But I still don't quite understand what this means.

Yes, he is talking about perpendicular component. But if I have a flat, infinite plane and the charge is uniformly distributed, perpendicular component is the only component, right?

Suppose that this charged plane lies in the x-y plane. At all points below z-axis the field is constant and negative, and at all points above z-axis, field is constant and positive, and there is discontinuity as we "cross" the boundary. Is the below graph correct?

If the field above is uniform and constant, and the field below is uniform and constant, then the difference is a simple matter of $E_{above}-E_{below}$. Instead, he applies Gauss' Law, states limiting condition where $\epsilon$ (distance between upper and lower Gaussian surface parallel to the plane) converges to zero, etc, etc... Intuition tells me that there must be something that I overlooked.

 

[jtbell]

I’m traveling now, so I don’t have my copy of Griffiths at hand to see the exact context for myself. However, that limiting process seems to be what one would use for the more general case of a non-planar sheet of charge, in which the field is not uniform at finite distances from the sheet.

Perhaps Griffiths is actually discussing the more general case, or perhaps he is simply anticipating it.

 

[Chandra Prayaga]

jtbell is right. Griffiths is indeed discussing the general case of a non-flat sheet of charge. The OP is a particular case. The diagram given by OP is correct for an infinite plane sheet of charge. But the relation in Griffiths is more generally valid. For an infinite flat sheet, as hypercube says, the perpendicular component is the only component. In the general case, even if there is a parallel com[ponent, the discontinuity in the perpendicular component still follows the relation given in Griffiths.

 

[Hypercube]

I understand now. Thank you for your replies.