Critical distance value for attractive force

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chaos333
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On Problem 3.11 for Griffiths' Electrodynamics, there is a question that asks for the critical value between a point charge and a conducting shell, but I don't quite know what they mean by 'critical value' in this context and how I'm supposed to approach this question, the rest of the problem is reasonable.
Textbook problem 3.11: find the force between a point charge q and a nearby spherical conductor carrying total charge Q, showing the force is attractive when separation a is less than a critical value a_c, and asking what that critical value is. Answer given: R(1 + sqrt(5))/2.
 
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The critical value would seem to be the point at which the force direction changes.
I think we will need to see Problem 3.10 to understand the set-up.

AM
 
Andrew Mason said:
The critical value would seem to be the point at which the force direction changes.
I think we will need to see Problem 3.10 to understand the set-up.

AM
The method it refers to is that in 3.10 you were supposed to set q'+q''=0, where q' and q'' lie inside the conductor and represent the charge of the conductor, so for this problem you essentially do q'+q''=q where q'' lies on the origin of the sphere, q' lies a distance 'a' (inside the sphere still) and a third charge q is outside the sphere a distance 'b'. It's a modification of this problem.
Left: a grounded conducting sphere of radius R with V = 0, and a point charge q a distance a from its center. Right: the image charge configuration, showing charge q' at distance b from the center, charge q at distance a, and position vectors r, r' and r'' with angle theta.

Text box reading Problem 3.10: a question about a grounded conducting sphere with the addition of a second image charge, asking what charge to use, where to place it, and the force between a point charge q and a neutral conducting sphere.
 
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So in this problem, 3.11, the conducting sphere is a hollow conducting sphere that is not grounded containing charge q?

All charges will be on the surface so the sphere will have a surface charge distribution (for a given position of the point charge, q at a distance ##a## from the centre of the sphere) that gives all charges the same potential. Since ##\vec{F}=-\frac{dU}{dx}## you will need to find the point at which ##\frac{dF}{dx}=-\frac{d^2U}{dx^2}=0##

AM
 
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Andrew Mason said:
So in this problem, 3.11, the conducting sphere is a hollow conducting sphere that is not grounded containing charge q?

All charges will be on the surface so the sphere will have a surface charge distribution (for a given position of the point charge, q at a distance ##a## from the centre of the sphere) that gives all charges the same potential. Since ##\vec{F}=-\frac{dU}{dx}## you will need to find the point at which ##\frac{dF}{dx}=-\frac{d^2U}{dx^2}=0##

AM
Does this seem like the correct expression for F using superposition?
Handwritten red equation on dark background: F = 1/(4πε0) [q''/a² + q'/(a-b)²]
 
chaos333 said:
The method it refers to is that in 3.10 you were supposed to set q'+q''=0, where q' and q'' lie inside the conductor and represent the charge of the conductor, so for this problem you essentially do q'+q''=q where q'' lies on the origin of the sphere,
Yes

chaos333 said:
q' lies a distance 'a' (inside the sphere still)
Did you mean to say q' lies a distance 'b' from the center of the sphere?

chaos333 said:
and a third charge q is outside the sphere a distance 'b'.
Did you mean q is a distance 'a' from the center of the sphere?

chaos333 said:
Does this seem like the correct expression for F using superposition?
View attachment 348079
Yes, except you left out the charge q in your expression for F.
 
TSny said:
Yes


Did you mean to say q' lies a distance 'b' from the center of the sphere?


Did you mean q is a distance 'a' from the center of the sphere?


Yes, except you left out the charge q in your expression for F.
Yeah my bad,
Equation for force F = q/(4π ε₀) times (1/a² + R/a³ − R/(a(a − R²/a)²)), written as plain mathematical text with fractions.
so I differentiate this and set it to 0? Pretend 'a' is a variable, maybe x, then I differentiate it like a variable right?
 
chaos333 said:
Yeah my bad, View attachment 348098
Still missing a factor of q. Otherwise, looks good.

chaos333 said:
so I differentiate this and set it to 0? Pretend 'a' is a variable, maybe x, then I differentiate it like a variable right?
No need to differentiate. You're not looking for a "critical point" in the sense of a maximum or minimum of ##F##. The critical point in this problem is the value of ##a## for which the force switches from repulsive to attractive. What is the value of ##F## at this critical point?
 
TSny said:
Still missing a factor of q. Otherwise, looks good.


No need to differentiate. You're not looking for a "critical point" in the sense of a maximum or minimum of ##F##. The critical point in this problem is the value of ##a## for which the force switches from repulsive to attractive. What is the value of ##F## at this critical point?
0, that makes sense... When I go to solve for it though I just get a nasty polynomial, should I brute force it or is there another approach?
 
chaos333 said:
0, that makes sense... When I go to solve for it though I just get a nasty polynomial, should I brute force it or is there another approach?
Yes, it's messy. You might let ##x = a/R## or ##x = R/a##, whichever you prefer. As you're seeing, ##F = 0## yields a polynomial in ##x## for which you need to find the root(s). If you have access to a computer algebra system, you can check to see if the polynomial factors nicely. Or, you could plot the polynomial to get a good numerical approximation for the solution.
 
TSny said:
Yes, it's messy. You might let ##x = a/R## or ##x = R/a##, whichever you prefer. As you're seeing, ##F = 0## yields a polynomial in ##x## for which you need to find the root(s). If you have access to a computer algebra system, you can check to see if the polynomial factors nicely. Or, you could plot the polynomial to get a good numerical approximation for the solution.
Sorry for getting to you late, but I got it, thanks for your help!
 
chaos333 said:
Sorry for getting to you late, but I got it, thanks for your help!
You are welcome. Glad it worked out.