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If i have an infinite conductor plane near a point charge ( that is the configuration of the common "Method of images" example), if i calculate the conductor's inducted charged density as

$$\sigma =-\varepsilon _{0}\frac{\partial V}{\partial n}$$

that means that the electric field is only due to the conductor's inducted charged and not also due the point charge. Where am i wrong?

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No, it doesn't mean the electric field is only due to the conductor's induced charge. This is so because $V$ in your expression is the resulting potential due to the point charge and the induced ones on the plane. This resulting potential gives you information about all charges that produced it. Taking its laplacian on the location of the point charge gives you $\nabla^2V=-4\pi q\delta^3(\vec{r}) /\varepsilon_0$, so it tells you a point charge also produced $V$; and using the condition for the descontinuity of the potential's normal derivative on the condutor's surface gives you the induced surface charge $\sigma$, as in your expression, which is indeed one of the sources of such $V$.

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  • $\begingroup$ But dV/dn is the electric field perpendicular to the conductive plane, and this electric field is contemporary due to the point charge and the inducted charges. For this reason why should't i isolate the induced charge that, in addiction with the point charge, gives the dV/dn mentioned above ? If i use the formula wrote in the post, it seems that all the electric field perpendicular to the conductive plane is generated by the inducted charge... $\endgroup$ – Poiera Nov 23 '17 at 11:19
  • $\begingroup$ Remember where the formula of your post comes from. It actually tells you the difference of the potential's normal derivative equals to the surface charge. Therefore, the surface charge is responsible to a discontinuity on $E_\bot = -\partial V/\partial n$ instead of being responsible to $E_\bot$ itself. $\endgroup$ – andrehgomes Nov 23 '17 at 11:29
  • $\begingroup$ What kind of discontinuity ? Isn't dV/dn the conductor's perpendicular Electric field ? So dV/dn is the total electric field since near the conductor it only has the perpendicular component. Where am i wrong ? $\endgroup$ – Poiera Nov 23 '17 at 13:00
  • $\begingroup$ In particular, dV/dn is contemporary due to the induced charge and the point charge. But using the formula of the initial topic is like saying that all the dV/dn is only due to the induced charge $\endgroup$ – Poiera Nov 23 '17 at 13:47
  • $\begingroup$ Imagine there are two regions in your problem. One outside the condutor and another one inside it. The formula in your post comes from the condition $E_\bot^\text{outside}-E_\bot^\text{inside}=\sigma/\varepsilon_o$ evaluated at the interface between the two regions, i.e., at the conductor's surface. Because it is electrostatics, the electric field inside the conductor vanishes, remaining only $E_\bot^\text{outside}= -\partial V/\partial n=\sigma/\varepsilon_o$. Therefore, the function $E_\bot$ is discontinuous across the two regions because of the surface charge density. $\endgroup$ – andrehgomes Nov 23 '17 at 14:28

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