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Induced Charged on a Grounded Sphere

  • #1

Homework Statement:

A grounded conducting sphere of radius R_0 is centered at the origin. If we place a charge +q at z=3R_0, calculate the total induced charge Q on the sphere surface.

Relevant Equations:

\sigma = -\epsilon_{zero} dV/dn(R=R_0)
I've come to the result (using cylindrical coordinates)
#\sigma (z) = (-2q) / (pi*sqrt(R_0*(10R_0-6z)^3) )#
and i tried to get #Q# by integrating #2*pi*sqrt(R_0^2-z^2)*\sigma(z)dz# from #-R_0# to #R_0#.
But i can't solve that integral. I tried solving it numerically with arbitrary values and it didn't make sense.
I figured it should be independent of #R_0#, and we should come to# Q=-q/3#...
Any help please? Am i integrating it wrong? Or is it that the charge distribution i got might be wrong?
 
Last edited:

Answers and Replies

  • #2
sorry i nover posted before i thought if i wrote eqs in latex it would just come out right
 
  • #3
haruspex
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The latex needs to be enclosed in pairs of hash symbols.
Please post your working.

Homework Statement:: A grounded conducting sphere of radius ##R_0## is centered at the origin. If we place a charge +q at ##z=3R_0##, calculate the total induced charge Q on the sphere surface.
Relevant Equations:: ##\sigma = -\epsilon_{zero} dV/dn(R=R_0)##
I've come to the result (using cylindrical coordinates)
##\sigma (z) = \frac{-2q} {\pi\sqrt{R_0(10R_0-6z)^3} }##
and i tried to get Q by integrating ##2\pi\sqrt{R_0^2-z^2}\sigma(z)dz ## from -##R_0 ## to ##R_0.##
we should come to Q=-q/3...
 
  • #4
haruspex
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I'd go with spherical, not cylindrical, coordinates - if I tried to solve it via a coordinate system.
.
PS this is a trick question.
I wouldn't call it a trick question, but there is certainly an elegant solution. Almost a one-liner.
 
  • #5
haruspex
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I would call it a trick question when most of the given data is irrelevant, which it is.
Are you referring to the "relevant equation"? I see nothing else irrelevant.
 
  • #6
haruspex
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I'm referring to everything except. q.
Remember, the correct answer is given as -q/3 (with which I agree).
 

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