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Point charge above xy plane, find surface charge density!

  • Thread starter satchmo05
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Homework Statement


A point charge, Q, is located @ (0,0,d) above infinite conducting plane that lies in xy plane and is maintained at ground potential. Find:
a.) surface charge density as a function of x and y on conducting plane, and
b.) total charge induced on conducting plane.


Homework Equations


See below.


The Attempt at a Solution


Well, I know that the E field is upward directed in the z-direction because the E-field is always perpindicular to a conducting surface. The answer is in the back of the book, and it is:
a.) ρs = (-2Qd)/(4*pi*[ρ2 + d2](3/2)
b.) Qsurface = -Q

In part a, I do not see how the author got [ρ2 + d2](3/2) in the denominator of this formula. This equation shows up in the E-field equation for a uniform disc of charge, and how can we assume this?! If I were to follow this implementation, assuming I can follow the assumptions for an uniform disc of charge, r = daz, and r' = xax + yay. (r - r') = daz - xax - yay. The magnitude of this term to the third power (still following equations for uniform disc of charge) = sqrt(d2 + x2 + y2) = sqrt(d2 + ρ2).

I get stuck here. If someone could let me know if I should continue to follow this method, and if so, how to continue. If I should try another way of thinking, please offer some tips to get me going in the right direction. Thank you for all help, it's most appreciated!
 

Answers and Replies

  • #2
ehild
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  • #3
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can you explain how they got the potential formula and then how they got to the surface charge density on the grounded plane (the next equation)? is there a formula for this or something that I am missing here. From my text, I see no relationship that incorporates the surface charge density and the potential function, involving the permittivity of free space constant.

The closest thing I see is:
surface charge density = (epsilon)*V/d = (epsilon*E) = D, because D = surface charge density at conducting planes.
 
  • #4
ehild
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If you are not familiar with the electric potential yet, see this:

http://www.shef.ac.uk/physics/teaching/phy205/lecture_7.htm [Broken]

ehild
 
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  • #5
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Thanks ehild, I figured it out. Going with your image charge suggestion, I determined that I could find the surface charge distribution by using the formula En = ps/permittivity of free space constant. I solved for the electric potential difference (V) by creating a "dipole" moment with the image charge. Then E = V/d, and then used E to solve for ps, as stated earlier.

Thanks again.
 
  • #6
ehild
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Well done!

ehild
 

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