If I have an infinite plane charged sheet with a uniform charge density $\sigma$ and I want to know the electric field at a point $P$ at a distance $\vec r$ away from the sheet, how would I do that?
I approach it like this knowing that this is completely wrong:
I take that point $P$ and draw a Gaussian surface (a sphere in the image) that passes through that point. And since the charge enclosed by the surface is zero, I conclude that the net flux $\phi_E$ through this surface is also zero. And since the net flux $\phi_E$ is zero, the field $\vec E$ is also zero.
Which I know is wrong but I don’t know why?
Best Answer
Gauss' law is always true but not always useful; your example falls in the latter category. To infer the value of $\vec E$ from $\oint \vec E\cdot d\vec S$ you need a surface on which $\vert \vec E\vert $ is constant so that $$ \oint \vec E\cdot d\vec S= \oint \vert \vec E\vert \, dS = \vert \vec E\vert \oint dS = \vert \vec E\vert S \, . \tag{1} $$ $\vec E$ is not constant on your sphere, meaning you cannot use (1) and pull $\vert \vec E\vert$ out of the integral and recover $\vert\vec E\vert$ through $$ \vert \vec E\vert = \frac{q_{encl}}{4\pi\epsilon_0 S}\, . $$ In your specific example, this is why $\oint \vec E\cdot d\vec S=0$ even though $\vert \vec E\vert$ is never $0$ at any point on your Gaussian surface. The $0$ results from the geometry of $\vec E\cdot d\vec S$ everywhere on the sphere rather than $\vert \vec E\vert=0$.
To proceed you need to use a Gaussian pillbox with sides perpendicular to your sheet because, by symmetry, the field must also be perpendicular to your sheet. Thus $\vec E\cdot d\vec S$ for all sides of the pillbox is easy to compute. If your pillbox passes through the sheet, it will enclose non-zero charge and, using simple geometry, one easily shows that the flux through the back cap will add to the flux through the front cap and you can recover the usual result.