Electric field of ring

  • #1

Homework Statement


Determine both the location and the maximum magnitude Emax of the electric field along the axis of a uniformly charged ring. (Use epsilon_0 for ε0, Q, and a as necessary.)


Homework Equations


dE= k_edq/r^2costheta
costheta= x/r


The Attempt at a Solution


i think i have to take the derivative of some forumla
that looks like this:
(kex)/(a^2 + x^2)^3/2 all multiplied by Q

and set this to zero and find a max

i really have no idea how to do this
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
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Homework Statement


Determine both the location and the maximum magnitude Emax of the electric field along the axis of a uniformly charged ring. (Use epsilon_0 for ε0, Q, and a as necessary.)


Homework Equations


dE= k_edq/r^2costheta
costheta= x/r


The Attempt at a Solution


i think i have to take the derivative of some forumla
that looks like this:
(kex)/(a^2 + x^2)^3/2 all multiplied by Q

and set this to zero and find a max

i really have no idea how to do this

That looks about right.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html#c2

Maybe take the differential using the quotient rule?

(d(f(x)/g(x))/dx = ( f '(x)g(x) - f(x)g '(x) ) / g2(x)
 
Last edited:
  • #3
yes i applied the quotient rule
but it gets extremely confusing..i end up with an answer like (a+x)(a+x) =0

which im pretty sure is incorrect
 
  • #4
LowlyPion
Homework Helper
3,097
5
yes i applied the quotient rule
but it gets extremely confusing..i end up with an answer like (a+x)(a+x) =0

which im pretty sure is incorrect

I didn't work it out, but I noticed that the divisor goes away when you set it to 0, leaving you just

( f '(x)g(x) - f(x)g '(x) ) = 0

or when f '(x)g(x) = f(x)g '(x)
 
  • #5
so that would leave us with

3a+3x(a^2+x^2)^3/2(kex) - ke(a^2+x^2)^3/2 = 0

???
wat abt the Q in the original equation
im not very good at calculus so this is pretty tough for me
 
  • #6
LowlyPion
Homework Helper
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so that would leave us with

3a+3x(a^2+x^2)^3/2(kex) - ke(a^2+x^2)^3/2 = 0

???
wat abt the Q in the original equation
im not very good at calculus so this is pretty tough for me

Going by the link:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html#c2

That gives you the equation to begin with and basically there's all the constants invariant in x, so Q just gets carried along through as part of that constant. Let's call it C.

E = C*x/(x2 + a2)3/2

Since it seems to be a common factor I don't think it affects Xmax as it looks to cancel out when you set it to 0.
(Of course it does matter in evaluating |E|.)

Using the quotient rule where f(x) = C * x, and g(x) = (x2 + a2)3/2

So doesn't it look like (using the chain rule to determine g'(x)) and then setting to 0 that :

C*(a2 + x2)3/2 = 3*C*x2(a2 + x2)1/2 ?

If that's right then it looks like it simplifies to

3x2 = (a2 + x2)

or

x = (a*√2)/2 ?

Not sure if it's right, but it feels right. No time to double check it. You certainly should.
 
Last edited:
  • #7
you are a genius!
thanks for clearing that up

but now using the x value how do i find Emax?
 
  • #8
LowlyPion
Homework Helper
3,097
5
you are a genius!
thanks for clearing that up

but now using the x value how do i find Emax?

By substitution into your original equation, now that you know x in terms of a.
 

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