Electric field of a point charge above a plane conductor

[Buffu]

Lets say we have a plane conductor lying on the xy plane extending to infinity in all directions; we assign zero potential to this place .Now bring a point charge $Q$ at $h$ distance above the plane ... .

Now to solve this problem we find an easily soluble problem whose solution or a piece of it can be made to fit the given problem. Here the easy problem would be to place a point charge $-Q$ at a distance $h$ below the plane. The field in the upper half meets the all the requirement of the original problem.

The potential of the conductor is fixed but we have in the system a point charge at which the potetial approaches zero, we can regard the point charge as the limiting case of a small, spherical conductor on which the total charge $Q$ is fixed. For this mixed boundary condition - potentials given on some surfaces, total charge on other- a uniqueness theorem also holds.

Now to calculate the field, consider a point at a distance $r$ from the origin ... . So the z component of the field of $Q$ at this point is $-Qh/(r^2 + h^2)^{3/2}$, the mirror charge $-Q$ also the contributes a equal field, so the total electric field is $-2Qh/(r^2 + h^2)^{3/2}$. Therefore $\sigma = \dfrac{-Qh}{(2\pi)(r^2 + h^2)^{3/2}}$.

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 ?

 

[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.

 

[Buffu]

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.

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 ?

 

[haruspex]

the total electric field by the both the charges at a point r from the origin

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

 

[vanhees71]

Here's the complete solution for this problem using the image-charge method:

https://www.physicsforums.com/threa...ges-and-conducting-plane.909946/#post-5732140

 

[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 ?