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Distance between equipotential surfaces

  1. Sep 16, 2015 #1
    1. The problem statement, all variables and given/known data
    A non-conducting sphere (radius 11.3 cm) has uniform charge density ρ = 0.596 μC/m3. Find the distance, in meters, between equipotential surfaces V1 = 16.2 Volts and V2 = 42.3 volts. (Distance is always positive.)

    2. Relevant equations
    V=kq/r
    ρ=Q/V

    3. The attempt at a solution
    ρ=0.596*10-6 C/m3 = Q/V ⇒ (0.596*10-6)=Q/(4/3*π*(0.113m)3)
    ⇒Q=3.60*10-9

    for V1
    16.2=k(3.60*10-9)/r1
    ⇒r1≅0.5

    for V2
    42.3=k(3.60*10-9)/r2
    ⇒r2≅1.3

    ⇒r2-r1=0.807m → this answer is wrong. :oldconfused:
     
  2. jcsd
  3. Sep 16, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    Volts relative to what?

    There are missing units.
    How do you get a larger radius for the larger potential? Something got wrong with the multiplication/division there, you got the inverse values. An error that would have been easy to spot with units.
     
  4. Sep 16, 2015 #3
    OK thanks! I don't know how I messed up that algebra so bad. After re-checking my calculations this is what I got:

    r1 m = k N*m2/C2 * (3.6*10-9) C / 16.2 N*m/C

    r1 = 1.998 m

    r2 m = k N*m2/C2 * (3.6*10-9) C / 42.3 N*m/C

    r2=0.7651 m → smaller for the higher potential

    r1-r2 = 1.23 m

    Now it seems to make more sense. Thank you!
     
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