1. Not finding help here? Sign up for a free 30min 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!

Charged Ring again.

  1. Aug 30, 2009 #1
    1. The problem statement, all variables and given/known data

    problem.jpg

    2. Relevant equations

    [tex]E=\frac{kqz}{(z^2+r^2)^{3/2}}[/tex]

    3. The attempt at a solution

    (c) is asking where the maximum value of the electric field would be in terms of R. In order to do this, I have to take the derivative of this function, set it equal to zero, correct?

    [tex]E=kq\frac{z}{(z^2+r^2)^{3/2}}[/tex]

    Is this how I do this?

    [tex]\frac{dE}{dz} = \frac{uv\prime - u\prime v}{v^2}[/tex]
     
  2. jcsd
  3. Aug 30, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    Yes that is how to differentiate it.
     
  4. Aug 30, 2009 #3
    I end up with

    [edited]

    First of all, is this right?
     
    Last edited: Aug 30, 2009
  5. Aug 30, 2009 #4
    Okay, I'm completely lost on this one. I need help.
     
  6. Aug 31, 2009 #5

    rock.freak667

    User Avatar
    Homework Helper

    Where did your differentiation go?
     
  7. Aug 31, 2009 #6
    [tex]kq(\frac{(z)(3/2)(z^2+r^2)^{1/2}(2z)-(1)(z^2+r^2)^{3/2}}{(z^2+r^2)^3})[/tex]

    [tex]kq\frac{(3z^2)(z^2+r^2)^{1/2}-(z^2+r^2)^{3/2}}{(z^2+r^2)^3}[/tex]

    [tex](3z^2)(z^2+r^2)^{1/2}=(z^2+r^2)^{3/2}[/tex]

    [tex](3z^2)=(z^2+r^2)[/tex]

    [tex]2z^2-r^2=0[/tex]
     
  8. Aug 31, 2009 #7

    rock.freak667

    User Avatar
    Homework Helper

    so z= ± r/√2
     
  9. Aug 31, 2009 #8
    But what does that tell me in terms of the question asked?
     
  10. Aug 31, 2009 #9

    rock.freak667

    User Avatar
    Homework Helper

    so if Emax occurs for z=R/√2


    To find Emax, put z=r/√2 into your equation for E
     
  11. Aug 31, 2009 #10
    I inadvertently dropped the kq from the post above.

    [tex]
    kq(3z^2)(z^2+r^2)^{1/2}=(z^2+r^2)^{3/2}
    [/tex]

    [tex]
    kq(3z^2)=(z^2+r^2)
    [/tex]

    [tex]
    kq(2z^2-r^2)=0
    [/tex]

    [tex]z=\frac{r}{\sqrt{2kq}}[/tex]

    If I put it back into the original equation, I still have an unknown r then. This equation is a mess.
     
  12. Aug 31, 2009 #11

    rock.freak667

    User Avatar
    Homework Helper

    well you could just compute the value of z and then put that number into the equation with E
     
  13. Aug 31, 2009 #12
    Okay I can't solve this. The algebra is way too messy and I can't follow what I'm doing.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Charged Ring again.
  1. Charged Ring (Replies: 14)

  2. Ring of charge (Replies: 3)

  3. Charged Ring (Replies: 2)

  4. A ring of charge (Replies: 5)

Loading...