Physics olympiad - electrostatics

In summary: The first line is correct, the second only if you put in the right value for q. In summary, the answer key tells me that (a), (b), (c) are correct.
  • #1
alexmahone
304
0
(Any number of options may be correct.)

Three concentric spherical shells have radii r, 2r and 3r with charges q1, q2 and q3 respectively. Innermost and outermost shells are earthed. Then,

(a) q1+q3=-q2
(b) q1=-q2/4
(c) q3/q1=3
(d) q3/q2=-1/3

The answer key tells me that (a), (b), (c) are correct. Could anyone please help me solve this?
 
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  • #2
I know that I must write the net potentials of the innermost and the outermost shells as zero.

Could someone please tell me how to write the net potential of a particular shell? Does it depend on the other shells as well?

Thanks in advance.
 
  • #3
Show your work...which laws do you think you'll need?
 
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  • #4
borgwal said:
Show your work...which laws do you think you'll need?

[itex]\phi=\frac{1}{4\pi\epsilon}\frac{q}{r}[/itex]
 
  • #5
That equation is correct for a single point charge (or a charge q distributed in a spherically symmetric way) once you choose your potential equal to zero at infinity.

Do you know Gauss' law?

Do you know how to calculate a potential once you know the electric field?
 
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  • #6
borgwal said:
That equation is correct for a point charge (or a charge q distributed in a spherically symmetric way) once you choose your potential equal to zero at infinity.

Do you know Gauss' law?

Yes I know Gauss' law.
 
  • #7
borgwal said:
Do you know how to calculate a potential once you know the electric field?

E=-dV/dr ?
 
  • #8
Apply Gauss' law to find E, then use E to find the potentials: not by integrating from infinity, but starting at the origin r=0.
 
  • #9
alexmahone said:
E=-dV/dr ?

Yes, kind of: you need the inverse relation since you can calculate E from Gauss' law.
 
  • #10
borgwal said:
Yes, kind of: you need the inverse relation since you can calculate E from Gauss' law.

I'll try your method tomorrow and tell you if I was able to solve it or not. Anyway, thanks for your help. :)
 
  • #11
Okay: you should find that exactly *one* of the answers is correct.
 
  • #12
borgwal said:
Okay: you should find that exactly *one* of the answers is correct.

Actually, the answer key tells me that (a), (b), (c) are correct.
 
  • #13
One condition follows from the given that the potentials on shells 3 and 1 are the same. The other two are really the same conditions then, and follow if you insist on having a potential zero at infinity (I didn't think you need that, but okay...you do).

And, did you use the approach I suggested?
 
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  • #14
Yes, but now you have to get the potential from the fields, start integrating from either r=0, or r=\infty.
Actually, in this problem r is a constant for some reason, so use a different variable for the variable distance.
(I think you mean your answer only for the constant distances given in the problem, but that is not enough).
 
  • #15
Using Gauss' law I get

[itex]E_14 \pi r^2=\frac{q_1}{\epsilon}[/itex]
[itex]E_24 \pi (2r)^2=\frac{q_1+q_2}{\epsilon}[/itex]
[itex]E_34 \pi (3r)^2=\frac{q_1+q_2+q_3}{\epsilon}[/itex]
 
  • #16
borgwal said:
Yes, but now you have to get the potential from the fields, start integrating from either r=0, or r=\infty.
Actually, in this problem r is a constant for some reason, so use a different variable for the variable distance.
(I think you mean your answer only for the constant distances given in the problem, but that is not enough).

Could you please complete the solution for me? I'm really stuck! :(
 
  • #17
No, I'm not going to complete it for you.

For a variable distance r between R and 2R (I'm using R now for what is called r in the problem), the field is E(r)=q1/4\pi \epsilon r^2. You have to integrate this from R to 2R to get the potential difference between shells 2 and 1. Then you do the same for r between 2R and 3R: that gives you the potential difference between shells 3 and 1, which should be zero. That actually gives you one of the conditions a,b,c,d, but I won't tell you which one.

If you can't integrate, then you can't solve the problem.
 
  • #18
yeah..keep going
 
  • #19
[itex]E=\frac{q}{4\pi\epsilon R^2}[/itex]
[itex]E=-dV/dR[/itex]
[itex]V_2-V_1=\int^{2r}_{r}\frac{q}{4\pi\epsilon R^2} dR[/itex]
[itex]V_2-V_1=\frac{q}{4\pi\epsilon R} (r\ to\ 2r)[/itex]
[itex]V_2-V_1=\frac{q}{4\pi\epsilon}(1/r-1/2r) [/itex]
 
  • #20
Now you have to put in the correct values of q...and then you're basically there!
 
  • #21
And the final point is: how do you get the outer shell to be at the same potential as at infinity: the way to do that is to have a zero field outside the third shell, and that gives you another condition.
 
  • #22
alexmahone said:
[itex]V_2-V_1=\frac{q}{4\pi\epsilon}(1/r-1/2r) [/itex]
[itex]V_3-V_2=\frac{q}{4\pi\epsilon}(1/2r-1/3r) [/itex]

The first line is correct, the second only if you put in the right value for q.
 
  • #23
alexmahone said:
[itex]V_3-V_1=\frac{q}{4\pi\epsilon}(1/r-1/3r) [/itex]

So this is the wrong conclusion (see previous post)
 
  • #24
alexmahone said:
[itex]V_2-V_1=\frac{2q}{4\pi\epsilon}(1/r-1/2r) [/itex]
[itex]V_3-V_2=\frac{3q}{4\pi\epsilon}(1/2r-1/3r) [/itex]

It's not simply 2q and 3q, but some combinations of q1, q2, q3
 
  • #25
alexmahone said:
[itex]V_2-V_1=\frac{q_2}{4\pi\epsilon}(1/r-1/2r) [/itex]
[itex]V_3-V_2=\frac{q_3}{4\pi\epsilon}(1/2r-1/3r) [/itex]

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

V_3-V_2 is determined by q_1+q_2
 
  • #27
borgwal said:
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
q_3 matters only for the field outside the three shells
 
  • #29
alexmahone said:
[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
[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
alexmahone said:
[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.
 
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  • #32
[itex]V_3-V_1=\frac{q_1}{4\pi\epsilon r}+\frac{q_2}{4\pi\epsilon 6r}=0?[/itex]
 
  • #33
alexmahone said:
[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
 
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