Concentric spheres and electrostatic induction

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Homework Statement


There's three thick, conducting concentric spheres with radii R1, R2, R3 (R1 < R2 < R3) with charge Q1, Q2, Q3 respectively.
a.- The middle one is now wired to the ground. Find its net charge
b.- The internal and external spheres are now wired. Find the distribution of all the charges

Homework Equations



The Attempt at a Solution


I think that when you wire to conductors, they have the same potential, but I'm clueless on how to apply this (if it's what I'm supposed to use at all).
 

Answers and Replies

  • #2
LowlyPion
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Try constructing some Gaussian surfaces.

The total flux through a closed surface will be the charge enclosed won't it?
 
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Sure, but I also need to know the charge on each side of the sphere (internal and external) because they have a certain thickness. Besides, doesn't the fact that the middle sphere is wired to the ground (in a.-) or the fact that the internal and external spheres are connected (in b.-) change things?

I mean, for b, the surface would be 2 concentric spheres I believe, and for a... I don't even know what the surface would be.
 
  • #4
LowlyPion
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The thicknesses and charge on each doesn't really matter. Remember these are conducting spheres, so thickness doesn't mean a whole lot either.

For instance in a) what is the flux through a spherical surface between the middle sphere and the outer sphere?
 
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  • #5
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The thicknesses and charge on each doesn't really matter. Remember these are conducting spheres, so thickness doesn't mean a whole lot either.
For instance in a) what is the flux through a spherical surface between the middle sphere and the outer sphere?
It's the middle sphere's charge (whatever it is, that's what I have to find) + the inner sphere's charge (Q1).

The fact that it is connected to the ground affects the charge somehow, so the middle sphere's charge can't be Q2, but I can't really say how it affects it, other than the fact that the ground and the middle sphere have the same potential.
 
  • #6
LowlyPion
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It's the middle sphere's charge (whatever it is, that's what I have to find) + the inner sphere's charge (Q1).

The fact that it is connected to the ground affects the charge somehow, so the middle sphere's charge can't be Q2, but I can't really say how it affects it, other than the fact that the ground and the middle sphere have the same potential.

Since the middle sphere itself is grounded there will be no field off the outer surface since it is necessarily 0 right? Since by Gauss you know that this integral over the closed surface must be equal to what's inside and since that adds to 0, then ...
 
  • #7
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The charge of the inner surface of the middle sphere is -Q1 and the charge of the outer surface is 0?

I'm still not sure about what being grounded implies. If I understood properly, when a sphere grounded, the sphere's electric field outside it is 0? Why so?
 
  • #8
LowlyPion
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The charge of the inner surface of the middle sphere is -Q1 and the charge of the outer surface is 0?

I'm still not sure about what being grounded implies. If I understood properly, when a sphere grounded, the sphere's electric field outside it is 0? Why so?

That's what it looks like to me. What else will the outer surface of the sphere be? There is no field from the outer sphere after all.
 
  • #9
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But the answer says that the charge is: -Q1 - Q3·(R2/R3).

So I was thinking:
Let 1, 2, 3 be the inner, middle and exterior spheres. Let A, B be their inner and outer surfaces (i.e.: Q3B is the exterior sphere's outer surface charge). Qn is sphere 2's net charge.

Q3B = Q1 + Qn + Q3
So Q3A = Q3 - (Q1 + Qn + Q3) = - Q1 - Qn.
Therefore Q2B = -Q3A = Q1 + Qn.

Q1B = Q1 so Q2A = -Q1B = -Q1.
We check that Q2A + Q2B = -Q1 + Q1 + Qn = Qn.

But still no clue about Qn.
 

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