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Voltage dealing with two parallel plates

  1. Feb 6, 2006 #1
    Hey guys. I'm having trouble analyzing a part of this problem.

    Two conducting, infinite sheets of charge (with thickness a) are fixed perpendicular to the x-axis as shown. The sheet from x = -2a cm to x = -a has a net charge per unit area -s and the sheet from x = +a to x = +2a has a net charge per unit area +s. (Note: the notation Vx1 » x2 is equivalent to Vx2 - Vx1.)

    So basically the left plate is situated at -2a to -a, and the right play is situated at a to 2a. I know the electric field flows to the left.

    Knowing this I must answer these questions.

    1. the potential going from the right side of the right plate and approaching the plate is............
    a. <0
    b. =0
    c. >0

    I believe that it is zero because there isn't a net electric field at this point. Is that right?

    2. The potential going from the edge of the right plate through the right plate to the other side is.......
    I want to say its >0, but I don't know exactly why. Can anyone lead me in the right direction here?

    3. The potential going from the inside of the right plate to the inside of the left plate is.........
    I say it is <0 because the electric field is uniform on the inside. Also, since you are moving to the negative end of the field, potential is negative.

    Any help is appreciated. Thanks!
  2. jcsd
  3. Feb 6, 2006 #2


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    Staff: Mentor

    By +s and -s, does one mean +[itex]\sigma[/itex] and -[itex]\sigma[/itex], which means the plate on the right (+x) has a positive charge and the plate on the left (-x) has a negative charge. Therefore the plate on the right has a higher potential than the plate on the left.

    The electric field points for + potential (+ charge) to - potential (- charge), by convention.

    Somewhere in between the potential must be zero, but the electric field is not zero anywhere between the plates. Think of a parallel plate capacitor.
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