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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 $(x,y)$?

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    $\begingroup$ Once you ground the sphere, no matter what vharge you give it, its always at potential 0. $\endgroup$
    – Lelouch
    Commented Aug 13, 2016 at 16:37
  • $\begingroup$ @Lelouch What is the reason for that? $\endgroup$
    – Alex
    Commented Aug 13, 2016 at 17:12

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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.

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  • $\begingroup$ But if we have a grounded conducting plane and a charge $q$ a distance $d$ above the plane as I described, then the charge $q$ induces a negative charge on the conducting sphere in a region closest to the charge. Hence in this region we can assume that $\rho \neq 0$ (since it is negative). Hence by Poisson's equation this implies that $\nabla^2 V = -\frac{\rho}{\epsilon_0}$ in this region. But if $V = 0$ since the plane is grounded, then we would get $\nabla^2 V = 0$ everywhere including the induced region. Which one is correct? Thanks. $\endgroup$
    – Alex
    Commented Aug 14, 2016 at 17:12
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    $\begingroup$ $V=0$ does not imply $\Delta V =0$! Think e.g. about the centre of a parabola $f(x)=x²$ which has $f(0)=0$ and $f''(0)=2$. $\endgroup$
    – Sanya
    Commented Aug 15, 2016 at 9:57
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    $\begingroup$ I think you are right. So we can still have $\nabla^2 V = -\frac{\rho}{\epsilon_0}$ where $\rho \neq 0$ and $V = 0$ on the plane. $\endgroup$
    – Alex
    Commented Aug 15, 2016 at 10:39

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