1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A ring with charge Q and radius 'r'

  1. Sep 15, 2014 #1

    squelch

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data

    Show that ##\vec{E}_x## on the axis of a ring charge [I'm assuming they meant "of charge Q"] of radius "r" has its maximum value at ##x=\pm \frac{r}{\sqrt{2}}##

    2. Relevant equations

    Linear charge density ##\lambda=\frac{Q}{2\pi R}##
    ##dQ=\lambda ds = \frac{Qd\theta}{2\pi}##
    ##\vec{E}=\frac{KQ}{R^2}##


    3. The attempt at a solution

    Let the distance between an infinitesimal sector of the ring with charge ##dQ## and a point ##P## be ##a##. Then ##a=\sqrt{x^2+r^2}## where r is the radius of the ring. Note that to save some typing, I will use ##a## and ##\sqrt{x^2+r^2}## interchangeably.

    Let ##\theta## be the angle formed by a and the x axis, then:
    ##cos\theta=\frac{x}{a}=\frac{x}{\sqrt{x^2+r^2}}## and ##sin\theta=\frac{r}{a}=\frac{r}{\sqrt{x^2+r^2}}##

    Now, ##d\vec{E}_y=0##, because all forces in the y-component cancel, leaving ##d\vec{E}_{total}=d\vec{E}_x=d\vec{E}cos\theta##

    ##\therefore d\vec{E}=\frac{KdQ}{a^2} cos\theta##

    Because ##dQ= \frac{Qd\theta}{2\pi}## and ##cos\theta=\frac{x}{\sqrt{x^2+r^2}}##
    ##d\vec{E}=\frac{KQx}{2\pi a^3}d\theta##

    Now, integrating from 0→2∏ to find ##\vec{E}##
    $$\vec{E}=\frac{KQx}{2\pi a^3} \int_{0}^{2\pi} d\theta = \frac{KQx}{a^3} = \frac{KQx}{(x^2+r^2)^{3/2}}$$

    From here, I'm not sure how to use this to "prove" that ##\vec{E}_x## has a maximum value at ##x=\pm \frac{r}{\sqrt{2}}##
     
  2. jcsd
  3. Sep 15, 2014 #2

    gneill

    User Avatar

    Staff: Mentor

    How do you go about finding the maximum or minimum of a function of x?
     
  4. Sep 15, 2014 #3

    squelch

    User Avatar
    Gold Member

    First derivative test?
     
  5. Sep 15, 2014 #4

    gneill

    User Avatar

    Staff: Mentor

    You've sort of got the right idea. Set the first derivative w.r.t. x to zero and solve for x. Then test the second derivative at x to see if you've found a maximum or a minimum. It's a Calculus 1 topic.
     
  6. Sep 15, 2014 #5

    squelch

    User Avatar
    Gold Member

    Right, but I don't feel like I have my ##\vec{E}## right. I keep seeing that ##\sqrt{x^2+r^2}## and thinking that I missed that I was supposed to do trig substitution somewhere along the way.
     
  7. Sep 15, 2014 #6

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You formula for E is correct. You need to find x where the function ##\frac{x}{(x^2+r^2)^{3/2}}## has its extrema.

    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: A ring with charge Q and radius 'r'
  1. Charged Ring (Replies: 2)

  2. A ring of charge (Replies: 5)

Loading...