Critical distance value for attractive force

  • #1
chaos333
11
1
Thread moved from the technical forums to the schoolwork forums
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.
1720558106482.png
 
Physics news on Phys.org
  • #2
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
 
  • #3
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.
1720560422732.png

1720561446850.png
 
Last edited:
  • #4
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
 
Last edited:
  • #5
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?
1720573200818.png
 
  • #6
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.
 
  • #7
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,
1720576226038.png
so I differentiate this and set it to 0? Pretend 'a' is a variable, maybe x, then I differentiate it like a variable right?
 
  • #8
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?
 
  • #9
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?
 
  • #10
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.
 
  • #11
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!
 
  • #12
chaos333 said:
Sorry for getting to you late, but I got it, thanks for your help!
You are welcome. Glad it worked out.
 

Similar threads

Replies
4
Views
137
Replies
1
Views
686
Replies
2
Views
1K
Replies
11
Views
1K
Replies
14
Views
2K
  • Electromagnetism
Replies
3
Views
1K
Replies
4
Views
997
Replies
2
Views
859
Replies
1
Views
1K
  • Electromagnetism
Replies
4
Views
939
Back
Top