Physics olympiad - electrostatics

  • #26
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V_2-V-1 is determined by the charge q_1,

V_3-V_2 is determined by q_1+q_2
 
  • #27
66
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V_2-V-1 is determined by the charge q_1,

V_3-V_2 is determined by q_1+q_2
Where does q_3 figure then?
 
  • #28
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q_3 matters only for the field outside the three shells
 
  • #29
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[itex]V_2-V_1=\frac{q_1}{4\pi\epsilon}(1/r-1/2r) [/itex]
[itex]V_3-V_2=\frac{q_1+q_2}{4\pi\epsilon}(1/2r-1/3r) [/itex]
Yes! Now set V_3-V_1 equal to zero...
 
  • #30
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[itex]V_2-V_1=\frac{q_1}{4\pi\epsilon}(1/r-1/2r)[/itex]
[itex]V_3-V_2=\frac{q_1+q_2}{4\pi\epsilon}(1/2r-1/3r)[/itex]

So I add these equations?
 
  • #31
367
0
[itex]V_2-V_1=\frac{q_1}{4\pi\epsilon}(1/r-1/2r)[/itex]
[itex]V_3-V_2=\frac{q_1+q_2}{4\pi\epsilon}(1/2r-1/3r)[/itex]

So I add these equations?
Sorry, yes! You need V_3-V_1, and then set the result equal to zero.
 
Last edited:
  • #32
66
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[itex]V_3-V_1=\frac{q_1}{4\pi\epsilon r}+\frac{q_2}{4\pi\epsilon 6r}=0?[/itex]
 
  • #33
367
0
[itex]V_3-V_1=\frac{q_1}{4\pi\epsilon r}+\frac{q_2}{4\pi\epsilon 6r}=0?[/itex]
Not quite....you made a small mistake in the addition
 
Last edited:

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