Electric Field of a Uniformly Charged Ring: Finding Emax Along the Axis

[physics_geek]

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 formula
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

 

[LowlyPion]

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 formula
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)

 

[physics_geek]

yes i applied the quotient rule
but it gets extremely confusing..i end up with an answer like (a+x)(a+x) =0

which I am pretty sure is incorrect

 

[LowlyPion]

yes i applied the quotient rule
but it gets extremely confusing..i end up with an answer like (a+x)(a+x) =0

which I am 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)

 

[physics_geek]

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

 

[LowlyPion]

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.

 

[physics_geek]

you are a genius!
thanks for clearing that up

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

 

[LowlyPion]

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.