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Finding Voltage at a point

  1. Feb 23, 2014 #1
    I want to find the potential at point P due to a finite, constant linear charge density λ, of length L.


    zVcpV14.png


    first, if i wanted to find the x-component of the E-field at point P..

    $$E=\int{\frac{k(λdx)}{r^2}}cosθ$$

    $$r=\sqrt{z^2+x^2}$$

    $$cosθ=\frac{x}{\sqrt{z^2+x^2}}$$

    $$E=kλ\int{\frac{1}{z^2+x^2}}\frac{x}{\sqrt{z^2+x^2}}dx$$

    evaluated from 0 to L yields..

    $$-kλ(\frac{1}{z^2+L^2}-\frac{1}{\sqrt{z^2}})$$

    --------------------------

    when we did found the potential at piont P in class, we did the following though

    $$V_{p}= -\int{\vec{E}\cdot d\vec{r}cosθ}$$

    my confusion is, when we add the cosθ term, aren't we just finding the voltage of the x-component E-field?? Because, i thought we were trying to find the total voltage at point P, but this suggests otherwise doesn't it?

    also..

    expanding $$V_{p}$$..

    since $$r=\sqrt{z^2+x^2}$$

    $$dr=\frac{1}{2}(z^2+x^2)^{-1/2}(2xdx)$$



    $$V_{p}= \int{\frac{k(λdx)}{z^2+x^2}\frac{x}{\sqrt{z^2+x^2}}(\frac{1}{2}(z^2+x^2)^{-1/2}(2xdx))=kλ\int{\frac{x^2}{(z^2+x^2)^2}}dx}$$

    (evaluated from 0 to L)

    is this expression for the potential correct? in class we got..

    $$-kλ\int{\frac{zx}{z^2\sqrt{z^2+x^2}}\frac{x^2}{z^2+x^2}dx}$$
     
    Last edited: Feb 24, 2014
  2. jcsd
  3. Feb 23, 2014 #2

    rude man

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    Why mess with E fields at all? Use dV = k dq/r etc. and integrate.

    Your last integral does not make sense ...
     
  4. Feb 24, 2014 #3
    i added the dx..

    other than that does it make sense? moreover is it the correct expression for the voltage at point P?

    using dV=kq/r i get..

    $$V=\int{\frac{k}{r}dq}=kλ\int{\frac{1}{\sqrt{x^2+z^2}}dx}$$

    but this is equal to neither of the two equations... so now i have three equations.. which one is correct??
     
    Last edited: Feb 24, 2014
  5. Feb 24, 2014 #4

    rude man

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    I won't deal with the E field. Much too laborious.


    Anyway, your equation here is where I'd put my money. Integrated from 0 to L.
     
    Last edited: Feb 24, 2014
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