I Electric field of a point charge above a plane conductor

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1. May 16, 2017

Buffu

Some things I did not get are,

1): In the third para, it is said that potential approaches $\infty$ at $Q$, why ?

2): In last para, The z component of field by $Q$ is said to be $-Qh/(r^2 + h^2)^{3/2}$, Why ? should not it be $+Q$ not $-Q$ has the charge itself is positive.

3): I know $E_n = 4\pi\sigma$, where $E_n$ is field due to the conductor near its surface and perpendicular to the surface, but here, in last para, the field $E_z$ is induced by the charge $Q$ and $-Q$ not the plane conductor. So how do we get $\sigma$ from this ?

4): Lastly I don't get the need of placing $-Q$ below the surface, why we did that ?

2. May 17, 2017

haruspex

It's rather garbled, but I think I understand most of it.
The third paragraph I cannot comprehend, but I believe you can largely ignore it. The relevant part is the reference to a uniqueness theorem.

We wish to find the charge distribution induced on an infinite plane conductor at z=0 by a point charge Q at (0,0,h).
The potential at the plane will everywhere be zero. If we throw the plane conductor away and instead have a point charge -Q at (0,0,-z) we will also get zero potential everywhere at z=0. The uniqueness theorem says the field at z≥0 is the same in both cases.

So we can calculate the z component of the field that would be generated at z=0 by the -Q and equate that to the z component of the field that is generated by the induced charge on the plane conductor. A point at distance r from the origin is considered.

3. May 17, 2017

Buffu

So as far I get, here a summary . We place $-Q$ charge at $-h$ to simulate the zero potential surface at xy plane. Then we found the total electric field by the both the charges at a point $r$ from the origin. Now if we place the conductor back and kick the added charge then this would be the field at the point $r$ from the origin.

Since we know that $E_{\perp}$ field at a point is $E_{\perp} = 4 \pi \sigma$ and the field we calculated equals $E_{\perp}$, hence was the further calculations.

Am I correct ?

4. May 17, 2017

haruspex

No, just the lower one. The field from the conducting plane must be the same as the field from its -Q replacement.

5. May 18, 2017

vanhees71

6. May 18, 2017

Buffu

Yes thanks I finally understand this.
Can I still do this with a finite plane as long as the point is lying on the plane ?