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Help with partial derivative electric field

  1. Oct 27, 2008 #1
    We found on-axis potential of a ring of radius R and Charge Q to be:
    V=(1/4pi*Epsilon naught) * (Q/sqrt(z^2 + R^2))


    Find on axis electric field of ring of charge

    I know i just derive that equation, but am getting stuck.

    i got d/dz((1/4pi*Epsilon naught) * (Q/sqrt(z^2 + R^2))

    Any ideas how i do that partial derivative?
     
  2. jcsd
  3. Oct 27, 2008 #2

    Dick

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    Use the chain rule. d/dz(f(z)^(-1/2))=(-1/2)*(f(z)^(-3/2))*d/dz(f(z)). Same as always.
     
  4. Oct 27, 2008 #3

    Mark44

    Staff: Mentor

    If I understand your function, it's a function of z alone, so you don't need to take a partial derivative.

    d/dz((1/4pi*Epsilon naught) * (Q/sqrt(z^2 + R^2))
    = Q*(1/4pi*Epsilon naught) *d/dz[ z^2 + R^2]^(-1/2)

    Now, just use the chain rule to differential (z^2 + R^2)^(-1/2)

    I'm assuming that Q, R, and [tex]\epsilon[/tex]0 are constants.

    Also, it's not clear whether the expression with [tex]\epsilon_{0} [/tex] is [tex]\frac{1}{4pi * \epsilon_{0}}[/tex] or [tex]\frac{1}{4pi} \epsilon_{0}[/tex]. The use of parentheses will make it clear.
     
  5. Oct 27, 2008 #4
    so z becomes (-1/2)*(f(z)^(-3/2))*d/dz(f(z))? i understand that but how does that then become:

    E=(Q/(4pi*Epsilon naught)) (zQ/(z^2 + R^2)^3/2?

    thanks!
     
  6. Oct 28, 2008 #5

    Dick

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    In your best interests, I'm not going to tell you. Work it out. What do you get? Show us how you did it if you still have questions. BTW I think E is the negative of the gradient of the potential.
     
  7. Oct 28, 2008 #6

    Mark44

    Staff: Mentor

    No, z doesn't become what you have above; that's dV/dz. Also, you started with V as a function of z, so you should end up with dV/dz, not E.
     
  8. Oct 28, 2008 #7

    Dick

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    Hi Mark44, It's actually a physics problem. V is voltage and E is electric field. The gradient of V is E (up to a sign). And it is a partial derivative. The d/dx and d/dy parts are zero because V is cylindrically symmetric and krugertown is working on the z-axis.
     
  9. Oct 28, 2008 #8

    Mark44

    Staff: Mentor

    My mistake. I was thinking that E was EMF.
     
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