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Equipotential surface problem

  1. Nov 24, 2015 #1
    1. The problem statement, all variables and given/known data
    d.png


    2. Relevant equations
    ##\vec{E}##=##\frac{∂V}{∂r}##

    3. The attempt at a solution
    I have provided both problem and solution(almost)but the problem is I did not understand the solution.First of all
    I did not understand the question what we are told to determine?
    I locate the charges as follows
    GRAPH.png
     
  2. jcsd
  3. Nov 24, 2015 #2

    andrewkirk

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    Did you write the red part? It seems pretty sensible to me, so if you wrote it I think you've finished.

    If it's a supplied solution and you don't understand it, can you explain what it is that troubles you about it? That would make it easier to help.

    I don't know if this is relevant to your concern but I'll point out anyway that the surface they've asked you to identify is the only flat equipotential surface. All the others curve away from the flat surface. So the problem in general isn't as trivial as it might seem (if it's the apparent triviality that worries you).
     
  4. Nov 24, 2015 #3
    Yes.
     
  5. Nov 25, 2015 #4
    I mean the question mentioned two charges located at rest and then suddenly it (the question)points out at equipotential surface.I don't understand from where does this equipotential surface come?
     
    Last edited: Nov 25, 2015
  6. Nov 25, 2015 #5

    haruspex

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    Do you understand what an equipotential surface is?
     
  7. Nov 25, 2015 #6
    The flux is radiating radially outward for positive charges and inward if negative. For stand alone charge, any circle around it will be equipotential. If you introduce another charges, every point is determine by the resultant potential.
     
  8. Nov 25, 2015 #7
    Equipotential surfaces are surfaces of constant scalar potential.
     
  9. Nov 25, 2015 #8

    haruspex

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    Right. So, in the configuration of charges given, you are trying to find the locus of all the points at some particular potential. In general that locus will be a surface, but it is here given as a 2D problem, so you are just looking for a closed curve in the plane. You are told a point that is on it (which effectively determines what the constant potential is).
    Suppose the point (x, y) is on the equipotential. Write out the expression for the potential at that point.
     
  10. Nov 25, 2015 #9
    es.png
    Right?
     
  11. Nov 25, 2015 #10

    haruspex

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  12. Nov 25, 2015 #11
    How?
     
  13. Nov 25, 2015 #12
    Potential of the equipotential surface shown(by me in post #9)
    ##V##=##\frac{q}{4πεo r}##

    Potential due to +1.0 C=##\frac{1}{4πεo 0.5}##

    Potential due to -1.0 C=##\frac{-1}{4πεo 0.5}##

    Net ##V## at the point =0

    Hence this potential is constant for the entire surface
    Right?
     
    Last edited: Nov 25, 2015
  14. Nov 25, 2015 #13
    equipotential surface(especially when it is 2d) is just made up of many points at which potential is of same value.Right?
     
  15. Nov 25, 2015 #14
    What I don't understand is we are given only two points how can we determine that the entire surface should be represented by straight line?beyond those two points ,surface can take any route/shape.
     
    Last edited: Nov 25, 2015
  16. Nov 25, 2015 #15
    I think this somehow answers my question.But problem is I did not understand what @andrewkirk has written.
     
  17. Nov 25, 2015 #16

    Doc Al

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    You've calculated the potential at a single point.

    The trick is to realize that, due to symmetry, the potential at any point on the plane y = 1.5 will be zero.
     
  18. Nov 25, 2015 #17

    andrewkirk

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    Each equipotential surface is a surface in 3D space, which corresponds to a line on the 2D graph, defined by the equation 'potential = P'. Each value of P gives a different line. The line they have asked you to find is the one where P=0.

    The potential at a point ##(x,y)## is calculated as the sum of the potentials at that point created by each of the two charges, ie
    ##\frac{1}{4\pi\epsilon_0}\bigg(\frac{1}{d((x,y)(1,1))}-\frac{1}{d((x,y),(1,2))}\bigg)## where ##d((x,y),(u,v))## is the distance between points ##(x,y)## and ##(u,v)##.

    For P=0, this is a straight line that bisects the two charges. For P>0 it will be a downwards curved line, reflectively symmetric in the line x=1, with its maximum at x=1 at a point closer to the lower charge [Or it may be the other way around. I haven't bothered to work out which way the signs go because it doesn't matter for the purpose of this exercise] For any given potential value P, the equipotential line will be the line satisfying the equation

    ##\frac{1}{4\pi\epsilon_0}\bigg(\frac{1}{d((x,y)(1,1))}-\frac{1}{d((x,y),(1,2))}\bigg)=P##
     
  19. Nov 25, 2015 #18
    How do you know?
    What's that?
     
  20. Nov 25, 2015 #19

    haruspex

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    You would not need to be given any points if you were just given the potential, V. You could then find all the points where the potential is equal to that value.
    At the point (x, y), what is the potential due to the charge at (1, 1)? What is the potential due to the charge at (1,2)? Write the equation that the sum equals V.
    Instead of being given the value V, you are given two points where the value is V. That allows you to determine V.
     
  21. Nov 25, 2015 #20
    But I nowhere wrote that we have been given potential V
     
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