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Homework Help: What is the relationship between electric field strength and the potential gradient?

  1. Feb 10, 2010 #1
    1. The problem statement, all variables and given/known data

    What is the relationship between electric field strength and the potential gradient?



    2. Relevant equations



    3. The attempt at a solution

    This is my Calc based Physcis lab question but I am at a total loss. I do not understand what potential gradient is in the first place. From google I figured out that it is rate of change.

    My understanding is as the gradient decreases the electric field strength increases. But I know that by looking at graph I have made. I do not understand why is it so theoritically.

    Please help!

    Thanks,

    Arshad
     
  2. jcsd
  3. Feb 10, 2010 #2
    Re: What is the relationship between electric field strength and the potential gradie

    http://en.wikipedia.org/wiki/Potential_gradient

    Check that out, thinking about it more mathematically;

    You agree that the electric potential is given by [tex] v = \frac{kq1q2}{r^2} [/tex] , right?

    Now when looking for the potential gradient, you're looking for the change of potential per distance moved away, so if for instance you plotted a graph of electric potential against distance away, the potential gradient is basically the gradient of that graph, so [tex] \frac{dV}{dr} [/tex] .

    Now find [tex] \frac{dV}{dr} [/tex] where [tex] V = \frac{kq1q2}{r^2} [/tex]

    you might see a similarity with E.
     
  4. Feb 11, 2010 #3
    Re: What is the relationship between electric field strength and the potential gradie

    THanks! :) This makes much more sense to me now! :)

    So basically, the more steeper the gradient (or equipotential lines) is, the mroe stronger the electric field gets, right?
     
  5. Feb 11, 2010 #4

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    Re: What is the relationship between electric field strength and the potential gradie

    Actually that's not quite right.

    [tex] V = - \int _P E \cdot dl, [/tex]

    Where P is an arbitrary path (i.e. V is the "voltage" between the two endpoints of the path),

    Not that it matters much for this problem, but the potential of a point charge, with respect to infinity, is:

    [tex]
    V = -\int _{\infty} ^{r} \frac{1}{4 \pi \epsilon _{0}} \frac{q}{r' ^{2}} dr' [/tex]

    [tex] = \frac{1}{4 \pi \epsilon _{0}} \frac{q}{r} [/tex]

    Note that it is a function of 1/r, not 1/r2.

    But none of that really matters for this problem.

    I think the question is asking you to look up a formula in your book, and compare it to the gradient of the potential.

    The gradient of a scalar field is specified by the [tex]\nabla[/tex] operator (called the "del" operator). So the gradient of the potential is specified by [tex] \nabla V [/tex].

    So look in your textbook for something that relates [tex] \nabla V [/tex] to [tex] E [/tex]. Although I haven't seen your textbook, I'm confident this relationship is in there.
     
  6. Feb 12, 2010 #5
    Re: What is the relationship between electric field strength and the potential gradie

    Well, there is one:

    E = -dV/dl

    OR

    V = - integral [ E dl]

    Thanks soo much to both of you for yours' help!! :)
     
  7. Feb 12, 2010 #6
    Re: What is the relationship between electric field strength and the potential gradie

    Yeah >.< sorry Arshad, replace the 1/r^2 with a 1/r in my post, foolish on my part.
     
  8. Feb 20, 2012 #7
    Re: What is the relationship between electric field strength and the potential gradie

    Is any one here to explain the Result of electric field as potential gradient
    E= -dV/dr
    Please Explain this result
    what is affect on electric intenity as Electric potential difference increase or Decrease
    And aslo what is affect on electric intenity distance r increase or Decrease
    Please Help me?
     
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