# Equipotentiality of the conductor surface with extra charges sprinkled and Poisson's equation

The electrostatic potential $$\phi(x,y,z)$$ everywhere on the surface of a conductor/metal is the same. For a quick reference see this post. Hence, any derivative of $$\phi(x,y,z)$$ must be zero on the surface. Now assume that we sprinkle extra charges on the surface of a metal (e.g., by rubbing it with silk). Does the metal surface still remain an equipotential surface? If so, how can the equation $$\nabla^2\phi=-\rho/\varepsilon_0$$ hold (Lhs zreo, Rhs nonzero)?

## 2 Answers

Actually, it is not true that any(!) derivative $$\phi(x,y,z)$$ must be zero on the surface. If $$\vec{t}$$ is the tangent vector to the surface, $$\nabla \phi \cdot \vec{t} =0$$ under the assumption that the surface is smooth which is by far not always the case. However, the derivative $$\nabla \phi \cdot \vec{n}$$ can be very well non-zero where $$\vec{n}$$ is the normal vector on the surface. Therefore there is no contradiction with Poisson's equation.

• Could you please explain in your answer whether throwing extra charges on the surface still leaves the conductor surface to be an equipotential one? Say, I rubbed the surface with silk. – mithusengupta123 Jun 5 at 13:09
• If additional charges are put on the perfect conductor's surface they will distribute in a way that the potential on the surface will be constant, however, this constant potential value might be different from initial one. – Frederic Thomas Jun 5 at 14:19

You have to be careful with the usage of derivative when multiple variables are involved. It is true that if you were to look at the potential along some curve that is on the conductor's surface then you would in fact find no change in potential. In other words, if you were to look at your potential as a function of some parameter $$s$$ that tells you how far along the curve on the conductor's surface you are, you would indeed see a constant function whose derivative with respect to $$s$$ is $$0$$.

However, we are not bound to the conductor's surface for all derivatives. You could look at the change in potential as you move normal to the surface, or even at some other angle. There are infinitely many directions you could look at. Therefore, just saying "the derivative is $$0$$" is not specific enough. There isn't a single derivative of your potential function.

In Possion's equation, the term $$\nabla^2\phi$$ is not a derivative in one direction. In Cartesian coordinates it is equal to $$\nabla^2\phi=\frac{\partial^2\phi}{\partial^2x}+\frac{\partial^2\phi}{\partial^2y}+\frac{\partial^2\phi}{\partial^2z}$$

which depends on multiple dimensions in general.

• So even if I choose $x,y$ directions to be along the surface, say of a sphere at a point, I have $\partial_x\phi=\partial_y\phi=0$ but not $\partial_z\phi$. In other words, on the surface of a sphere, in general, $\partial_r\phi\neq 0$. – mithusengupta123 Jun 5 at 13:15
• @mithusengupta123 Yes, that is correct. Or to make it even simpler just think about a sheet of charge – Aaron Stevens Jun 5 at 13:43
• Thanks. Can you explain why the equipotentiality, in the sense I you wrote, still holds when I rub the surface with a piece of cloth and transfer unneutralized charges? – mithusengupta123 Jun 5 at 13:47
• @mithusengupta123 charges on conductors move in such a way so that the field is $0$ throughout the entire conductor. Since $$\Delta V=\int\mathbf E\cdot\text d\mathbf l$$ this means the potential cannot be changing as you move around the conductor – Aaron Stevens Jun 5 at 14:32