What is the Maximum Electric Field of a Ring?

[Stang289GT]

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

1) dE= (ke)dq/r^2cos(theta)
cos(theta)= x/r

2) Ex=Q*((ke*(x))/(a^2 + x^2)^3/2)
(a is the radius of the ring)

The Attempt at a Solution

I took the derivative of Equation 2 and set it equal to 0 to find the maximum x value
x=(a*sqrt(2))/2 and I know that to be the correct answer for where the maximum E occurs; but upon attempting to plug this back into the original equation to find what the maximum E would be, I get the wrong answer every time. Any advice as to what I'm doing wrong?

Any help would be greatly appreciated.

 

[kuruman]

Hi Stang289GT. Welcome to PF.

If you do not show what you have done, I cannot figure out where you went wrong. Also, please note that sometimes when you write something up for someone else to see, you find your own mistakes.

 

[Stang289GT]

Hmmm...okay.
(there's a figure that goes along with this, but I can't upload it...)

Plugging in the maximum x=(a*sqrt(2))/2 value into equation 2 you get:

Emax=(k_e*Q*((a*sqrt(2))/2))/(((a*sqrt(2))/2)^2 + a^2)^(3/2)
simplifying this, I think it's supposed to be:
Emax=(k_e*Q*sqrt(2))/6a
but this answer is not correct. Is there an equation other than this one that I should be using?

 

[kuruman]

Assuming that your expression for the E-field is correct, your problem is probably in your simplification. It would help if you wrote

$$\frac{a \sqrt{2}}{2}=\frac{a}{\sqrt{2}}$$

then the denominator becomes

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

and simplify from there.