# Homework Help: Electric Field Due to a Line of Charge Problem

1. May 24, 2007

### frankfjf

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

Charge is uniformly distributed around a ring of radius R = 2.40cm and the resulting electric field magnitude E is measured along the ring's central axis (perpendicular to the plane of the ring). At what distance from the ring's center is E maximum?

2. Relevant equations

Therein lies my problem, I'm unsure how to approach this except ink knowing that Calculus can handle it. I know that I need to find the maximum and can obtain it by deriving the equation, finding the value of Z, then plugging it in and deriving a second time. But the only equation I know of here is:

E = kqz/(z^2 + R^2)^3/2

3. The attempt at a solution

I'm not sure if I have to modify the formula before I derive it or if I only worry about one part of it to derive. If I just derive the formula in terms of z I get a huge mess. What am I missing? I'm not sure where to start.

2. May 24, 2007

### Staff: Mentor

By "derive" I assume you mean differentiate. Go ahead--take the derivative! It's a bit messy, but so what? (Use the quotient/product rule.)

3. May 24, 2007

### frankfjf

Well after taking the derivative and using the quotient and chain rules I get something like...

kq * (z^2 + R^2)^3/2 - z(3/2(2z^3+2R^3)^1/2) / (z^2 + R^2)^5/2

But I get the feeling I either did something wrong in the differentiation or I can still further simplify.

4. May 24, 2007

### robphy

Try factoring.

5. May 24, 2007

### frankfjf

I can see that working but I can't figure out how to best factor it out, should I favor the entire top half or just the right side of the top half?

6. May 24, 2007

### Staff: Mentor

Looks to me like you made a few errors. For one, how did you get the term that I highlighted?

I recommend that you use the product & chain rules, writing the field equation as:

E = kq [z] [(z^2 + R^2)^-3/2]

(I put brackets around the two factors of interest, in case we need to refer to them; the kq drops out--since you'll set it to zero.)

7. May 24, 2007

### frankfjf

Oh, I had forgotten about that approach. In that case once I've applied both the product and chain rules I end up with..

-3z(z^3 + R^3)^-5/2 + (z^2 + R^2)^-3/2

8. May 24, 2007

### Staff: Mentor

I spoke too soon. What's up with the highlighted term? Where'd you get the cubes?

Let's do this step by step: What's the derivative of the second bracketed factor in post #6?

Last edited: May 24, 2007
9. May 24, 2007

### frankfjf

Chain rule gives me... -3/2(z^2 + R^2)^-1/2 * (2z + 2R), hence how I get the cubes.

10. May 24, 2007

### Staff: Mentor

Forget simplifying until you fix it. Answer my question in my last post.

11. May 24, 2007

### frankfjf

Sorry, I replied too soon, I've editted post 9 to answer your question.

12. May 24, 2007

### Staff: Mentor

Several problems here:
(1) The derivative of (z^2 + R^2) is just 2z. (R is a constant!)
(2) Where did you get the power of -1/2? Remember you started with a power of -3/2.
(3) Even with the wrong derivative, how do you get cubes from that?

Last edited: May 24, 2007
13. May 24, 2007

### frankfjf

Ahh good point, d'oh. Actually, chalk it up to temporary insanity, doing it a second time I get...

(z^2 + R^2)^-3/2 + (2z^2)^-3/2

14. May 24, 2007

### Staff: Mentor

Not sure what this is. One more time, redo the derivative of the second factor in post #6.

15. May 24, 2007

### frankfjf

Trying it again I get -3/2(2z)^-5/2

16. May 24, 2007

### Staff: Mentor

You need to review the chain rule. Do it in two steps:
(1) Take care of the exponent part: (something)^-3/2
(2) Take the derivative of the something

(And then multiply them, of course.)

17. May 24, 2007

### frankfjf

-3(z^3 + R^2z)^-5/2 ?

18. May 24, 2007

### Staff: Mentor

Nope. Do each step listed in my last post separately. (Don't try to combine anything.)

19. May 24, 2007

### frankfjf

I need some clarification then. By "second factor" do you mean what would be considered g'(x) in the product rule?

IE: f'(x)g(x) + g'(x)f(x)?

If so then g'(x), in steps, looks like...

(1) (-3/2)(z^2 + R^2)^-5/2
(2) 2z

When I multiply them, I get -3(z^3 + zR^2)^-5/2, unless I am doing something incorrectly, but I don't think I am, since step 1 would also require the power rule.

But this is identical to my previous answer, so I'm not sure what I'm doing wrong but it seems to be in step 1. But it seems to line up just fine with the power rule. The exponent is -3/2, which when I subtract 1 becomes -5/2, and then the original exponent is multiplied by the inside of the parenthesis.

Last edited: May 24, 2007
20. May 24, 2007

### Staff: Mentor

Exactly correct!

You are doing something wrong when you multiply:
z*(something)^-5/2 does not equal (z*something)^-5/2 !

(Don't waste time multiplying this out correctly--that will only make it harder to simplify the final expression.)

21. May 24, 2007

### frankfjf

So I should leave the final expression for g'(x) as 2z(-3/2)(z^2 + R^2)^-5/2

and continue from there?

22. May 24, 2007

### Staff: Mentor

Yes. (But please cancel the 2! )

23. May 24, 2007

### frankfjf

So -3z(z^2 + R^2)+^-5/2?

24. May 24, 2007

### Staff: Mentor

Yes. (Except for that + sign.) Now put it all together.

25. May 24, 2007

### frankfjf

Oh, I apologize, that was a typo.

Since we've got f'(x)g(x) + f(x)g'(x), wouldn't that be...

(z^2 + R^2)^-3/2 - 3z^2(z^2 + R^2)^-5/2?