Electric Field Due to a Line of Charge Problem

In summary: 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) 2zIn summary, at a distance of 2.40cm from the ring's center, the electric field magnitude is maximum.
  • #1
frankfjf
168
0

Homework Statement



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?

Homework 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

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.
 
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  • #2
frankfjf said:
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.
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
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
Try factoring.
 
  • #5
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
frankfjf said:
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.
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
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
frankfjf said:
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
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?
 
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  • #9
Chain rule gives me... -3/2(z^2 + R^2)^-1/2 * (2z + 2R), hence how I get the cubes.
 
  • #10
frankfjf said:
I'm not sure how to simplify that though.
Forget simplifying until you fix it. Answer my question in my last post.
 
  • #11
Sorry, I replied too soon, I've editted post 9 to answer your question.
 
  • #12
frankfjf said:
Chain rule gives me... -3/2(z^2 + R^2)^-1/2 * (2z + 2R), hence how I get the cubes.
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?
 
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  • #13
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
frankfjf said:
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
Not sure what this is. One more time, redo the derivative of the second factor in post #6.
 
  • #15
Trying it again I get -3/2(2z)^-5/2
 
  • #16
frankfjf said:
Trying it again I get -3/2(2z)^-5/2
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
-3(z^3 + R^2z)^-5/2 ?
 
  • #18
frankfjf said:
-3(z^3 + R^2z)^-5/2 ?
Nope. Do each step listed in my last post separately. (Don't try to combine anything.)
 
  • #19
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.
 
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  • #20
frankfjf said:
If so then g'(x), in steps, looks like...

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

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

and continue from there?
Yes. (But please cancel the 2! :wink:)
 
  • #23
So -3z(z^2 + R^2)+^-5/2?
 
  • #24
frankfjf said:
So -3z(z^2 + R^2)+^-5/2?
Yes. (Except for that + sign.) Now put it all together.
 
  • #25
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?
 
  • #26
Looks good to me. Now simplify. (Eliminate the common factor in both terms)
 
  • #27
That's where it gets tricky. I'm not sure if I can eliminate z^2 + R^2 since that leaves 1^-3/2 and 1^-5/2 and mathematics life isn't usually that easy, so... I guess the only common term I can think of is z^2 itself since it appears 3 times? That'd give...

z^2[(1 + R^2)^-3/2 - 3(1 + R^2)^-5/2?
 
  • #28
frankfjf said:
That's where it gets tricky. I'm not sure if I can eliminate z^2 + R^2 since that leaves 1^-3/2 and 1^-5/2 and mathematics life isn't usually that easy, so... I guess the only common term I can think of is z^2 itself since it appears 3 times? That'd give...

z^2[(1 + R^2)^-3/2 - 3(1 + R^2)^-5/2?
Yikes no! You cannot just pull things out from under a power!

(z^2 + R^2)^-3/2 is not equal to z^2*(1 + R^2)^-3/2

(z^2 + R^2)^-3/2 is not equal to (z^2 + R^2)*1^-3/2

You need to review your understanding of exponents and powers.

Hint: (z^2 + R^2)^-3/2 itself is a common factor--divide it out from all terms.
 
  • #29
Ahhhh okay. Sorry, been a while since I dealt with messy expressions. However, for me the tricky part of factoring (z^2 + R^2)^-3/2 out of the second term. I'll try though... I get...

(z^2 + R^2)^-3/2 [1 - 3z^2(z^2 + R^2)^-1] I'm not sure if this is correct.
 
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  • #30
Looks good--you corrected your typo.

Keep going.
 
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  • #31
Hmm well to be honest I'm kind of stumped again, I'm not sure if there's anything left to factor out... couldn't I just set equal to zero and attempt to find the value for z that would cause that to happen, or is there further simplification to be done?
 
  • #32
Well, I think I can re-arrange the equation if I can't simplify further to be...

(1-3z^2) / (z^2 + R^2)^3/2 in which case z must equal the square root of 1/3?
 
  • #33
frankfjf said:
Hmm well to be honest I'm kind of stumped again, I'm not sure if there's anything left to factor out... couldn't I just set equal to zero and attempt to find the value for z that would cause that to happen, or is there further simplification to be done?
Just set it equal to zero and then simplify.

frankfjf said:
Well, I think I can re-arrange the equation if I can't simplify further to be...

(1-3z^2) / (z^2 + R^2)^3/2 in which case z must equal the square root of 1/3?
Not sure how you got this.
 
  • #34
I fear I might need a hint again. If I put an "= 0" at the end, I'm not sure how I'd get something on the other side. Only thing I can think of is to use the binomial theorem on the expression in the brackets, but I'm not sure if such a thing is fair play or helps to simplify it.
 
  • #35
Take the derivative (your expression in post #29) and set it equal to zero. Realize that each factor can be set equal to zero, if possible. (Don't dare use the binomial theorem.)
 

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