Field due to ring charge at x-y axis.

[Buffu]

Homework Statement

A charge $Q$ is distributed unifromly around a thin ring of radius $b$ which lies in $xy$ plane with its centre at the origin. Locate the point on the positive $z$ axis where the electric field is strongest ?

Homework Equations

The Attempt at a Solution

$\displaystyle d\vec E = {dQ \over b^2 + r^2 }\cos \alpha$. $r$ is the distance between the point, $\alpha$ is the angle between $z$ axis and the displacement vector , and the ring along $z$ axis.

$\displaystyle d\vec E = {\lambda dl \over b^2 + r^2 } \cos \alpha = d\vec E = {\lambda bd\theta \over b^2 + r^2 } \cos \alpha$.

Using $\displaystyle \cos \alpha = {r \over \sqrt{b^2 + r^2}}$.

$\displaystyle E = {br\lambda \over (b^2 + r^2)^{3/2} } \int^{2\pi}_{0} d\theta = {2\pi br\lambda \over (b^2 + r^2)^{3/2} }$

Since $\displaystyle \lambda = {Q\over 2\pi b}$,

I get,

$\displaystyle E = {Qr \over (b^2 + r^2)^{3/2}}$

Which has a maximum $\displaystyle r = {2\over 3\sqrt{3}b^2}$.

Am I correct ?

 

[kuruman]

The derivation of the electric field is correct. Your final answer is not. If b is the radius, your value for r has the wrong dimensions. It must proportional to b not to b^-2.

 

[Buffu]

The derivation of the electric field is correct. Your final answer is not. If b is the radius, your value for r has the wrong dimensions. It must proportional to b not to b^-2.

Oh that was WA's mistake. I got lazy and let it calculate maxima. Is $r = \pm b/ \sqrt{2}$ correct ?

 

[kuruman]

Where is b on the right side of the equation?

 

[Buffu]

Where is b on the right side of the equation?

sorry.

I edited my post. Is the answer $\pm b/\sqrt{2}$ ?

 

[kuruman]

I would pick the positive value because the problem specifies "on the positive z axis". Otherwise it's fine.