casseverhart13 20190906 05:46:42  Thank you a bunch for sharing this with all people you actually know what you are talking approximately! Bookmarked. house painting
danielsw98667 20191021 06:33:35 
Well, with the application of Gauss law the answer should be D. Villa Las Rocas

 
gigadan 20130929 13:10:15  How about we treat it as a dipole with d distance separation?
Then we know for sure answer has to contain qd
Also, the field of a dipole goes off by r^3
gigadan 20130929 13:12:39 
I meant separation of 2d
So.. the 4pi would become 2pi as a result.

 
walczyk 20121107 15:57:22  The official solution is very good except that it starts off with the wrong sign. The term with (zd) is the original charge, and (z+d) is the mirror charge [remember, if you increase z you should move closer to the original charge], and we also are given that the particle has charge q. Therefore the problem is missing a factor of 1, which explains the negative charge density given in the solution.  
mpdude8 20120419 15:06:22  Dimensional analysis is your best friend on a 100 question exam.  
nakib 20100402 08:40:10  For grounded conductor .
By symmetry of the problem [remember that you have an opposite 'image' charge at d],
Figure out . Choice (D) is the answer.
But one need not go through all these if (s)he notes that only (D) has the right units.  
kroner 20091005 23:28:58  The conducting plate shields the far side from the electric field of the point charge so the field there is zero. On the near side close to the plate the parallel components are zero, and the perpendicular component should be double what's contributed by the point charge alone. So the field on the near side is
where is the angle from the perpendicular so .
Applying Gauss's Law on a volume that contains a small area of the plate,
.
 
proctort 20091005 12:49:16  What is " is the direction of the directest straight line" supposed to mean?rnIs "directest" even a word?  
irishroogie 20071004 18:31:37  As erc pointed out, choice D is the only one with the correct units of charge over area.
ramparts 20091105 15:20:12 
You can do this the hard way, or you can do this the smart way. Unfortunately, yosun's answers, while a huge help, are often far more complicated than they need be, and not particularly amenable to a 100question multiplechoice test.

 
RootMeanSquare 20061125 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...)  
eigenrosen 20061027 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 20081010 19:26:00 
Yeah the answer to this question definitely needs to be rewritten!

 
eigenrosen 20061027 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.  
l3amm 20061019 20:02:59  Alternatively, the only answer with the correct units is D. Simple, but little tricks can make hard problems really easy.  
erc 20051105 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.
Plantis 20100406 11:49:19 
Absolutly correct! I agree.

shak 20100731 13:15:29 
Compare answers with their units...all are eliminated except D which has unit of charge over area

 