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B Help Please -- E/M conceptual question regarding electric potential

  1. Jan 16, 2017 #1
    So in a graph where V is a function of x, when the slope is negative what does that mean about the direction of the field along the x axis? What about when the slope is positive?
    SerPSE8-25-p-034.gif
     
  2. jcsd
  3. Jan 16, 2017 #2

    berkeman

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    Welcome to the PF.

    Can you think of a charge distribution that might cause that V(x) graph? You might also consider if conductors might be involved... :smile:
     
  4. Jan 16, 2017 #3
    Would that be a hollow sphere?
     
  5. Jan 16, 2017 #4

    berkeman

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    That's not what I'm picturing, but perhaps it might work as well. But it would take multiple hollow concentric conducting spheres charged to different voltages, no?
     
  6. Jan 16, 2017 #5

    Doc Al

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    How does the field relate to the potential difference? Say you had a uniform field point to the right. Imagine placing a positive test charge in that field. Which way will the field push the charge? Will a positive test charge tend to move towards higher or lower potential?
     
  7. Jan 16, 2017 #6
    towards lower potential
     
  8. Jan 16, 2017 #7

    berkeman

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    Can you sketch the concentric conducting shells example that you alluded to? Can you label the shells with their voltages to make that graph work? :smile:
     
  9. Jan 16, 2017 #8

    Doc Al

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    Exactly. So what does that tell you about your diagram?
     
  10. Jan 16, 2017 #9
    ohhh so if the graph shows a positive slope then the test charge would be moving towards the positive charge and if its negative then it would be moving away?
     
  11. Jan 16, 2017 #10

    Doc Al

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    Where the graph shows a positive slope, the positive test charge would tend to move towards lower potential (to the left). So what direction must the field be in that region?

    So how does the field direction depend on the slope of the potential graph?
     
  12. Jan 16, 2017 #11
    How do you define field direction? Wouldn't saying left and right be arbitrary?
     
  13. Jan 16, 2017 #12

    Doc Al

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    Just specify the direction with respect to the x-axis given. ("To the right" corresponds to the +x direction, in this case.)
     
  14. Jan 16, 2017 #13
    ohhhhh. So in that case, the region defined by the positive slope is probably to the left of a positive test charge. The area in which there is a constant V value is at the charge and then the negative slope would be to the right of the test charge because there is lower PE the larger the x value.
     
  15. Jan 16, 2017 #14
    its like when x is equal to 0(the reference point) that is some distance to the left of the pos charge right?
     
  16. Jan 16, 2017 #15

    Doc Al

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    What's the direction of the field?
     
  17. Jan 16, 2017 #16
    To the left?
     
  18. Jan 16, 2017 #17
    So does this mean that a positive slope means moving to the left and neg slope means moving to the right?
     
  19. Jan 16, 2017 #18

    Doc Al

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    Yes. Where the potential rises to the right, the field points left.
     
  20. Jan 16, 2017 #19
    THANK YOU SO MUCH FOR HELPING ME <3
     
  21. Jan 16, 2017 #20

    Doc Al

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    Yes. The field is related to the negative (opposite) of the slope of the potential graph. So where the slope is positive, the field is negative -- pointing to the -x axis (to the left, in this case). And where the slope is negative, the field is positive.

    You are very welcome.
     
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