Force due to a uniformly charged ring

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
7 replies · 9K views
Jimmy25
Messages
69
Reaction score
0

Homework Statement



A uniformly charged ring has a radius a, lies in a horizontal plane, and has a negative charge given by -Q. A small particle of mass m has a positive charge given by q. The small particle is located on the axis of the ring.

What is the minimum value of q/m such that the particle will be in equilibrium under the action of gravity and the electrostatic force? (Use the following variables as necessary: a, k, Q, and g.)?

Homework Equations



I let x = the distance the charge is from the center of the ring.

[tex]E[/tex]=[tex]\frac{kQx}{\sqrt{a^2+x^2}^{3}}[/tex]

The Attempt at a Solution



I started by saying that "mg" must be equal to:

[tex]\frac{kQxq}{\sqrt{a^2+x^2}^{3}}[/tex]

However, I cannot eliminate the x variable from the problem. Am I missing something or do I have the wrong approach altogether?
 
Physics news on Phys.org
You do not need to eliminate x. Express q/m as function of x from the condition qE=mg, and find where this function has its minimum.

ehild
 
I can express q/m as a function of x however it is also a function of several other variables which have no numerical values. How do I go about finding a minimum for the function algebraically?
 
"Use the following variables as necessary: a, k, Q, and g".

You need to give a formula, not a number.

ehild
 
Okay. But how do I find a formula for the minimum using only the variables a, k, Q, and g? I tried to take a derivative (thinking I may be able to then find a local minimum) but it got really nasty.

Am I missing something? - Another way to find the minimum?
 
[tex]mg = \frac{kQxq}{\sqrt{a^2+x^2}^{3}} \rightarrow q/m= \frac{g}{kQ}\frac{(x^2+a^2)^{3/2}}{x}[/tex]



Find the position of minimum of the function.

[tex]f(x)=\frac{(x^2+a^2)^{3/2}}{x}[/tex]

Replace back this x into the formula for q/m.

ehild
 
I got it! I had to grunt it out and it turns out the answer was not that pretty but it was correct nonetheless. Thanks for your help!
 
Congratulation! It was not easy to get the result, but you did it!