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Calculating Work and E-field from and equipotential contour graph

  1. Jul 12, 2011 #1
    1. The problem statement, all variables and given/known data

    prob04a_threeqcontour.gif

    A) Calculate the work performed by an external agent to move a charge of -0.59x10-12 C from `i' to `b'.

    B) Calculate the magnitude of the electric field at `k'

    2. Relevant equations

    W = q∆V

    Where ∆V is the potential difference between the two potential lines each point
    lies on

    E = -∆V/∆L

    3. The attempt at a solution

    A) I know the work equation is charge x the change in V but I am not sure how to interpret the Equipotential lines in order to get change in V.

    B) I know the equation is E = -∆V/∆L but I am not sure how to calculate both those variables from the graph.

    The real question I have is how to interpret this graph to get ∆V and ∆L? I know that an Equipotential contour graph shows the plain perpendicular to the e-filed thus making Volts zero at any place there isn't a point charge.
     
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  3. Jul 12, 2011 #2

    SammyS

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    The difference in elctric potential for two adjacent equi-potential "lines" is ±1 Volt.
     
  4. Jul 12, 2011 #3

    Redbelly98

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    (A)
    You have the right idea, you just need to find the potentials at points i and b in order to calculation ΔV

    For point i, notice that it is between the lines labelled 0V and +5. (Had you noticed the "+5" contour line?). How many lines over from the 0V line is i? How many volts is that?
     
  5. Jul 12, 2011 #4
    will that work ... i mean the lines are not symmetrical ... if they were we could say that pot. increase by +1 when you move every line ahead ... can we do that in this case also when lines are not symmetrically separated?
     
  6. Jul 12, 2011 #5
    Thank you tutors for your help that worked!
     
  7. Jul 12, 2011 #6

    Redbelly98

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    Yes, it will work. It is customary to draw contour lines with an equal increment in potential between adjacent contours. It's true for topographic maps as well, if you're familiar with those.

    You're welcome. :smile:
     
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