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Electrostatic Potential

  1. Apr 22, 2015 #1
    1. The problem statement, all variables and given/known data

    What is the electrostatic potential for the electric field in the region x ≥ 0 where:

    E
    = (E0)*e^(kxi) V/m, with E0 a constant?

    (The potential at x → infinity is defined to be zero).

    2. Relevant equations
    b
    v = v_a - v_b = - ∫ E . dl
    a


    3. The attempt at a solution

    0
    v = - ∫ (E0)*e^(kx) dx = (E0)/k.
    infinity

    Is this correct?

    Kind regards.
     
  2. jcsd
  3. Apr 22, 2015 #2

    BvU

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    Helo Alex,

    You can check for yourself if this is correct: how do you derive an electric field from a potential ? Does that work OK for your answer ?

    Oh, and did I miss a minus sign somewhere ?
     
  4. Apr 24, 2015 #3
    Thank you BvU !
     
  5. Apr 24, 2015 #4

    BvU

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    Does that mean you found a potential ##E_o\over k## V gives an electric field ## \vec E = E_0 \; e^{kx} \; {\bf \hat\imath} \ ## V/m ?
     
  6. Apr 24, 2015 #4

    BvU

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    Does that mean you found a potential ##E_o\over k## V gives an electric field ## \vec E = E_0 \; e^{kx} \; {\bf \hat\imath} \ ## V/m ?
     
  7. Apr 25, 2015 #5
    So I was given the electric field:

    ##\overrightarrow{E} = {E_0} e^{-kx}\hat i. ##

    Which is completely in the x-direction.

    Now using:

    ## V = {V_a} - {V_b} = - \int_a^b \overrightarrow{E}.\overrightarrow{dl}, ##

    with the above electric field we have:

    ## V = - \int_\infty^x {E_0} e^{-kx}dx = \frac{E_0} {k} e^{-kx}.##

    Then using what BvU suggested:
    I remembered from my notes that the electric field is related to the potential as follows:

    ##\overrightarrow{E} = - \nabla V.##

    So to test if the potential I found is correct, I just plug it in and find:

    ##- (\frac{\partial V} {\partial x})=-(-{E_0} e^{-kx}\hat i) ##

    Which gives us back our original Electric Field:

    ##\overrightarrow{E} = {E_0} e^{-kx}\hat i.##

    Thanks again BvU, much appreciated!
     
  8. Apr 25, 2015 #6

    BvU

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    well done !
     
  9. Apr 25, 2015 #7
    Initially my limits for integration were wrong, I had them from ##\infty## to 0.
     
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