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Charge density - parallel plates

  1. Oct 9, 2010 #1
    1. The problem statement, all variables and given/known data
    5. Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm.
    a) If the surface charge density for each plate has a magnitude of 47.0 nCm^-2, what is the magnitude of the electric field intensity in the region between the plates?
    b) What is the potential difference between the two plates? (2 marks)



    2. Relevant equations



    3. The attempt at a solution
    The wording here confuses me a bit, is Electric field intensity something different to the electric field?
    I've done E=4.7*10^-9/8.85E-12 =530.8N/C but is this the field intensity? Part b im not sure about, I need some help for it to.

    Thanks
     
  2. jcsd
  3. Oct 9, 2010 #2

    Doc Al

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    Staff: Mentor

    Electric field and electric field intensity: same thing. (Check your arithmetic: is the charge density 4.7 or 47?)

    For part b: You'll need to use the distance.
     
  4. Oct 9, 2010 #3
    47, so E=5310.7N/C
    Then V=kq/r but I dont know q so how would I do that?
     
  5. Oct 9, 2010 #4

    Doc Al

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    Staff: Mentor

    That formula is for a point charge, which is not relevant here. What's the general relationship between field and potential?
     
  6. Oct 9, 2010 #5
    found it, V=Ed

    so V=116.8V?
     
  7. Oct 9, 2010 #6

    Doc Al

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    Staff: Mentor

    You got it.
     
  8. Oct 9, 2010 #7
    Thanks
     
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