# Force due to a uniformly charged ring

1. Jan 15, 2010

### Jimmy25

1. The problem statement, all variables and given/known data

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.)?

2. Relevant equations

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

$$E$$=$$\frac{kQx}{\sqrt{a^2+x^2}^{3}}$$

3. The attempt at a solution

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

$$\frac{kQxq}{\sqrt{a^2+x^2}^{3}}$$

However, I cannot eliminate the x variable from the problem. Am I missing something or do I have the wrong approach altogether?

2. Jan 15, 2010

### ehild

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

3. Jan 15, 2010

### Jimmy25

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?

4. Jan 15, 2010

### ehild

"Use the following variables as necessary: a, k, Q, and g".

You need to give a formula, not a number.

ehild

5. Jan 15, 2010

### Jimmy25

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?

6. Jan 16, 2010

### ehild

$$mg = \frac{kQxq}{\sqrt{a^2+x^2}^{3}} \rightarrow q/m= \frac{g}{kQ}\frac{(x^2+a^2)^{3/2}}{x}$$

Find the position of minimum of the function.

$$f(x)=\frac{(x^2+a^2)^{3/2}}{x}$$

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

ehild

7. Jan 16, 2010

### Jimmy25

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!

8. Jan 16, 2010

### ehild

Congratulation! It was not easy to get the result, but you did it!