To find the electric field discontinuity $$\vec{E}_{above} – \vec{E}_{below } = \frac{\sigma}{\epsilon_0}\hat{n} \implies \frac{\partial V_{above}}{\partial n} – \frac{\partial V_{below}}{\partial n} = -\frac{\sigma}{ \epsilon_0}$$ across a boundary, we follow the approach as outlined in this post. In the derivation we took a very small and thin gaussian surface. Hence $\sigma$ is the surface charge corresponding to that patch of charged surface. For a conductor then, the field immediately outside is $$\vec{E} = \frac{\sigma}{\epsilon_0}\hat{n}$$ where $\sigma$ is still a surface charge of a patch.
Given a problem where we have a point charge $q$ held a distance $d$ above an infinite grounded conducting plane $(V = 0)$. The charge $q$ induces a certain amount of negative charge on the nearby surface of the conductor. We find the potential above the conductor to be $V(x,y,z)$. Then using the formula above, we have $$\sigma = -\epsilon_0\frac{\partial V_{above}}{\partial n}$$ since we have $V_{below} = 0$ and hence $\frac{\partial V_{below}}{\partial n} =0$. We eventually get $$\sigma(x,y) = \frac{-qd}{2 \pi (x^2 + y^2 + d^2)^{\frac{3}{2}}}.$$
Question:
Would the potential $V_{below}$ still necessarily be zero inside the conductor even after inducing a charge on the surface of the conductor? Is the surface charge $\sigma(x,y)$ interpreted as the surface charge of a small patch around every point $(x,y)$?
Best Answer
Grounded means, by definition, "having the same potential as infinity" and this is usually set to $0$ for convenience. As the whole conductor is an equipotential surface as any conductor, the potential underneath the conductor needs to be at the same potential if there are no charges underneath that could change that - because in a charge free region surrounded by a surface on which the potential is constant, the potential should remain constant too.
And yes, the surface charge $\sigma$ is the charge of a small infinitesimal part of the surface.