GREPhysics.NET: GR8677 #63

GR8677 #63

Problem

GREPhysics.NET Official Solution

Electromagnetism $\Rightarrow$ }Method of Image Charge

The boundary condition at the conducting plane is $V(0)=0$ . This doesn't mean that one can't put an ``image charge" a distance $-D$ away on the other side of the plane to make the calculation easier. Making the directest straight line from the charge to the plane along the $z$ axis, one gets the following image-charge potential:

$V = \frac{1}{4\pi\epsilon_0} \left(\frac{q}{\sqrt{x^2+y^2+(z-d)^2}}-\frac{q}{\sqrt{x^2+y^2+(z+d)^2}}\right).<br />$

The charge density for a grounded conducting plane is given by $\rho=-\epsilon_0 \frac{\partial V}{\partial n}$ , where $n$ is the direction of the directest straight line. The derivative is,

$\frac{\partial V}{\partial z}=\frac{1}{4\pi\epsilon_0} \left(\frac{q(z-d)}{\left(x^2+y^2+(z-d)^2\right)^1.5}-\frac{q(z+d)}{\left(x^2+y^2+(z+d)^2\right)^{1.5}}\right),<br />$

and thus, after combining terms, the density is

$-\frac{1}{2\pi}\frac{qd}{\left(x^2+y^2+(z-d)^2\right)^{1.5}}$

The problem wants the surface charge density a distance $D$ away from the point charge. So, plug in $\left(x^2+y^2+(z-d)^2\right)^{1.5}=(D^2)^{1.5}$ to get

$-\frac{1}{2\pi}\frac{qd}{D^3}$

Alternate Solutions

irishroogie
2007-10-04 18:31:37

As erc pointed out, choice D is the only one with the correct units of charge over area.

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Comments

irishroogie
2007-10-04 18:31:37

As erc pointed out, choice D is the only one with the correct units of charge over area.

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RootMeanSquare
2006-11-25 11:50:18

?? I don't quite get the setup of this problem. lets say the plane lies in the xy plane, the V(0) = V(Z=0) or what? And in which direction is D supposed to point?

Sorry if this very stupid, but I really don't get the problem (aside from the imension analysis...)

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eigenrosen
2006-10-27 12:22:42

And you can only combine the terms from the derivative after you plug in z=0 (the only place we would expect a surface charge).

tfw15
2008-10-10 19:26:00

Yeah the answer to this question definitely needs to be rewritten!

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eigenrosen
2006-10-27 12:12:25

I think the signs of the charges are off in the potential (given the assumption that the negative charge in the problem is in the +z direction, which you make when you take the normal derivative in the +z direction). This should account for the sign difference with the answer.

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l3amm
2006-10-19 20:02:59

Alternatively, the only answer with the correct units is D. Simple, but little tricks can make hard problems really easy.

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erc
2005-11-05 11:20:57

My reasoning was that (D) was the only one where the combination of d's and D's gave something with units of a surface density.

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