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Electrical field strength problem

  1. Feb 25, 2012 #1
    I have completed the force column in the table however im not sure what the question is asking me for the field strength calculation. am i to use the 25μC charge or the 8μC charge in the formula 1/4∏ε x Q/r^2, this may be obvious to someone with more experience of these questions,

    Thanks
     

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  3. Feb 25, 2012 #2

    gneill

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    Staff: Mentor

    Use the 8μC charge, as that's the charge producing the field that the 25μC charge experiences as per the wording of the problem.
     
  4. Feb 25, 2012 #3
    Thanks mate.
    I have calculated the force for the 1mm distance to be 1.8x10^6
    so for field strength i have used (1.8x10^6)/8μC = 2.25x10^11
    Does that answer look about right?
     
  5. Feb 25, 2012 #4

    gneill

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    Staff: Mentor

    Nope. You should have divided by 25μC thus leaving the 8μC charge in the calculation.
     
  6. Feb 26, 2012 #5
    For the table a graph is required, what values do you think they want? would it be force against field strength or would distance be used against another?
     
  7. Feb 26, 2012 #6

    gneill

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    Staff: Mentor

    At a guess I'd say two graphs would be appropriate: force vs distance and field strength vs distance. With clever axis labeling you might plot both on the same graph...
     
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