Critical distance value for attractive force

[chaos333]

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.

 

[Andrew Mason]

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

 

[chaos333]

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.

 

[Andrew Mason]

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

 

[chaos333]

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?

 

[TSny]

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

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?

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?

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.

 

[chaos333]

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,

so I differentiate this and set it to 0? Pretend 'a' is a variable, maybe x, then I differentiate it like a variable right?

 

[TSny]

Yeah my bad, View attachment 348098

Still missing a factor of q. Otherwise, looks good.

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?

 

[chaos333]

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?

 

[TSny]

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.

 

[chaos333]

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!

 

[TSny]

Sorry for getting to you late, but I got it, thanks for your help!

You are welcome. Glad it worked out.