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Mag of Electric Field at a point between 2 Parallel Plates

  1. Jan 31, 2014 #1
    1. The problem statement, all variables and given/known data
    The figure below shows two parallel plates. Assume they are infinite in area (this way we can neglect edge effects). The plate on the left is at a potential of +60V and the plate on the right is at a potential of 0V. The plates are a distance 20cm apart and the dashed lines are spaced equally 5cm apart.
    View attachment 66183
    *If the image doesn't work there is a positive plate on the left and a negative plate on the right. The distance between the 2 plates is 20cm. Point C is located 15 cm from the left plate and 5 cm from the right plate.*

    What is the magnitude of the electric field at point C?

    2. Relevant equations
    ΔV= - E * d * cosθ
    E = - (ΔV) / ( d * cosθ )

    3. The attempt at a solution

    I know ΔV=60 and cosθ= cos(0)= 1. I am given d=20 cm between the two plates.

    At first I was using d= 0.2, [E=-(60)/(.2*1)=300 N/C] but that answer is wrong, i think because the d=.2 value is for the whole system and not "Point C".

    So I thought that I should be using a d value that corresponds with the "Point C".

    What i am confused about is "Point C" is located at either the 5 cm or the 15 cm mark, depending which direction you are going. So it could either be...
    E = - (60) / ( .15 * 1 ) = 400 N/C
    E = - (60) / ( .05 * 1 ) = 1200 N/C

    However the website says both of these are wrong and I am not really sure why.

    Any help would be GREATLY appreciated!
  2. jcsd
  3. Jan 31, 2014 #2


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    Staff: Mentor

    The electric field has the same strength everywhere, so your first approach is correct (and you don't need the angular dependency there). The field strength does not depend on the position of C, as long as C is between the plates.

    If the electric field would be different for different points, none of your approaches would work.

    Maybe the website got confused with N/C? V/m is another way to express the field strength.
  4. Jan 31, 2014 #3
    Ok, the program required V/m not N/C, thank you for you're help!
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