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Show that potential difference can be defined in these two ways

  1. Apr 3, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that [itex]P.D = {|E|} × {d} [/itex] is equivalent to [itex] P.D = \frac{W}{Q}[/itex]

    Potential Difference equals electric field strength meters, or work done per unit charge

    (for the equation P.D = |E| X d I cannot remember if it is a dot or cross product. I'm asking this question from memory as I do not have the question sheet in front of me. Can someone please say if it is dot or cross and explain?)

    2. Relevant equations

    [itex]\underline{F} = \frac{qQ}{\underline{r}^2} * \frac{1}{4\pi\epsilon_{0}}[/itex] * [itex]\underline{\widehat{r}}_{qQ}[/itex]

    [itex]\underline{\widehat{r}}_{qQ}[/itex] Is the unit vector in the direction to get from q to Q


    [itex]\underline{E} = \frac{Q}{\underline{r}^2} * \frac{1}{4\pi\epsilon_{0}}[/itex] * [itex]\underline{\widehat{r}}[/itex]


    [itex]\underline{F} = \underline{E}q[/itex]


    [itex]V = - \oint\underline{E}.d\underline{l} [/itex]


    3. The attempt at a solution

    First I made sure that the units agree with each other.

    Basic units
    Length = m
    Mass = Kg
    Time = s
    Electric Current = A

    So...
    Force = m·kg·s-2
    Work done = m2·kg·s-2
    Electric field strength = m·kg·s-3·A-1
    Potential Difference = m2·kg·s-3·A-1
    Chrarge = s·A

    E meters will have units m2·kg·s-3·A-1
    Work done per unit charge will have units (m2·kg·s-2)/s·A = m2·kg·s-3·A-1
    Both of these are the units of electrical potential difference so the units agree.

    I then tried getting the equations to equal each other, or tried deriving one equation from the other. This is where I got stuck.

    |E|=q/(4(pi)epsilon0(r^2))
    |E|*r = q/(4(pi)epsilon0(r)

    W = ????? what equation should I use here?

    Thanks for reading!
     
    Last edited: Apr 3, 2013
  2. jcsd
  3. Apr 4, 2013 #2

    tiny-tim

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    Homework Helper

    Hi CraigH! :smile:
    (i think that should be E . d and W/I :wink:)

    you have to prove two equivalent definitions of voltage

    power/current = force "dot" distance …

    how is power related to energy?

    how is current related to charge?

    how is energy related to work done?

    how is work done related to force?

    how is field related to force and charge? :wink:
     
  4. Apr 4, 2013 #3

    BruceW

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    I think craig was using W to mean work, in which case, it should be W/Q. Also, I think when craig is writing P.D he means potential difference, i.e. electric potential difference (voltage). (Not power times distance).

    You've got the definition in your 'relevant equations' section:
    [tex]V = - \oint\underline{E}.d\underline{l}[/tex]
    There is a very similar equation relating force to work. If you can remember that, then you will be half-way to getting the answer.
     
  5. Apr 4, 2013 #4
    Hi Bruce,
    Yes I was using P.D for potential difference and W for work done. The equation for work done is:

    W= ∫F.dr

    The integral of force over distance. I believe this is also a dot product, as it is only the distance moved in the direction of the force that matters.

    So I'll try to use this to equate P.D=W/Q and P.D= ∫E.dl


    [itex]P.D = \frac{W}{Q}[/itex]

    [itex]P.D = \frac{\int F.dr}{Q}[/itex]

    [itex]F=EQ[/itex]

    [itex]E=\frac{F}{Q}[/itex]

    [itex]P.D = \int E.dr[/itex]

    Woo! I'm surprised I got this! Thank you!
     
  6. Apr 4, 2013 #5

    BruceW

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    Homework Helper

    Nice work man. This is such a nice analogy between (force and work) and (electric field and electric potential). Often if I get stuck on an electrodynamics problem, I think back to this analogy. But, you need to be careful, it can become more complicated if there are magnetic fields also.
     
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