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Gauss' Law - Planar Symmetry HELP!

  • Thread starter yanimated
  • Start date
  • #1
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Homework Statement


A square plate of edge length 9.0 cm and negligible thickness has a total charge of 6.3 x 10-6 C.

(a) Estimate the magnitude E of the electric field just off the center of the plate (at, say, a distance of 0.50 mm) by assuming that the charge is spread uniformly over the two faces of the plate.

(b) Estimate E at a distance of 62 m (large relative to the plate size) by assuming that the plate is a point charge.


Homework Equations



E = magnitude of electric field
Omega = Surface Charge Density
e = 8.85 x 10^-12 C^2/Nm^2
pi = pi (ie 3.14...)
r = radius
q = Charge

E = (Omega)/(2e)

and then I used E = [ 1/(4pie) ] [ q/r^2 ]

The Attempt at a Solution



Well, I first used the second equation there with q = 6.3 x 10^-6 C
and r = .5 mm = .0005 m to find the magnitude of the electric field.

And then I plugged what I got there into E = (Omega)/(2e) to find Omega.

That was wrong :(. What am I doing wrong? Is there another equation?
 

Answers and Replies

  • #2
mjsd
Homework Helper
726
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mmm... (a) is just the case of finding E field for an infiinitely large plate for the distance to plate is small compare to plate size... so r does not comes in at all...besides you had E = (Omega)/(2e), nowhere is there a "r" in here!

(b) as suggested by the hint, treat it as a point charge
 
  • #3
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"just off the center of the plate (at, say, a distance of 0.50 mm)"

So wouldn't the .50mm be the radius? Or am I just still not getting it?
 
  • #4
mjsd
Homework Helper
726
3
IF you consider that the plate is effectively an infinite plate, THEN you use E=(Omega)/2e as suggested by you... and there is no "r" in this equation!
 

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