How to Determine Charge Density on a Plane from a Point Charge?

[player1_1_1]

Homework Statement

hello, the problem is that i have charge $$q$$ which lies in $$d$$ distance from a plane and I need to find function $$\sigma$$ which describes density of this charge

The Attempt at a Solution

I wrote this situation in cylindrical coordinate system where $$q$$ lies in point $$A=(0,0,h)$$ and plane lies in $$z=0$$, because of symmetry I can suppose that this function will be $$\sigma=\sigma(\rho)$$, what now?
edit: there should have been $$\sigma=\sigma(\rho)$$ I don't know why edit don't work

 

[collinsmark]

I'm not quite sure I understand the problem statement. Is the plane conducting? Is it held at any particular potential (such as 0 V)?

If so, you might want to research what is called the "Method of image charges," or "Method of images." A quick Internet search should prove useful.

The idea is that you can replace the plane with an image charge (mirror image, mirror charge) at a distance d below where the plane was. Then solve for the electric potential of that system. Once you have that you can differentiate with respect to the height to find the electric field and evaluate it at an infinitesimal height above where the plane would be, then find the charge distribution that would cancel that electric field (at an infinitesimal height below the plane).

 

[player1_1_1]

thanks for answer, the plane is ground ($$V=0$$) and plane is conducting. Is it possible to do this with Maxwell's equations?

 

[collinsmark]

thanks for answer, the plane is ground ($$V=0$$) and plane is conducting. Is it possible to do this with Maxwell's equations?

I would imagine so, that yes you could. But that's not the approach that I personally would take.

• Replace the conducting plane with a point charge of -q at (0, 0, -d).
• Find the potential for all space. You have two point charges. This step shouldn't be too hard. The potential for one point charge is V = (1/[4πε₀])q/r. So just find the potential for the two point charges in terms of d, z and ρ instead of r.
• Find the electric field along the z-axis by finding E = -∂V/∂z, evaluated at z = 0.
• There is a simple relationship between σ and E for a charged plane. Use this relationship to solve for σ. (Hint: it's proportional to ε₀).

 

[player1_1_1]

thanks, its better, and the last question, how would this maxwell equation for this look like?

 

[collinsmark]

thanks, its better, and the last question, how would this maxwell equation for this look like?

I'm not really sure -- at least not directly. But we can break it up into pieces, and it would be the same thing I described before.

The system is in electrostatic equilibrium. We don't have to worry about any magnetism or changing currents or moving charges. So most of Maxwell's equations don't apply. But there is one that does, Gauss' law.

$$\oint _S \vec E \cdot \vec{dA} = \frac{Q_{enc}}{\epsilon_0}$$

This one is at least indirectly important for this problem. Using Gauss' law you could find the electric field for a single point charge if you wanted to. It doesn't say anything about the electrical potential, but it turns out that the electric potential for point charges is not really necessary to solve this problem, it just makes things easier. You could stick with the electric field of each of the two point charges from the beginning, and perform a vector sum. The reason I suggested using potential is because summing scalars is easier than vectors, even if there is a partial derivative involved later.

Then there is the relationship between σ and E for a charged plane. Gauss' law is what is used to derive that too. It's only tricky in this problem since the electric field inside the conductor is always zero, so you only need consider the field lines on the outside out of the conductor (and they are always perpendicular to the surface, just above the surface). But anyway, Gauss' law is what is used, is my point.

So when it comes down to it, the whole solution is embodied in Gauss' law, even if we use other concepts such as potential.