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Potential at the corner of a insulated charged cube

  1. Dec 5, 2011 #1
    1. The problem statement, all variables and given/known data
    An insulating cube of edge a has a uniform charge density p. The charge is zero everywhere outside the cube. The potential at an infinite distance from the cube is taken to be zero. If the potential at the center of the cube is Vo, find the potential at a corner of the cube. (Hint: Consider the potential at the center of a charged cube with the same charge density but with twice the length of the edge, use the principle of superposition in combination with a dimensional analysis.)

    The problem is, I don't know how to find the potential at the center of a charged cube with length of the edge 2a

    2. Relevant equations
    i don't know if i am able to use gauss law to find the E-field out side and inside the cube


    3. The attempt at a solution
    i know that i can make a cube of side 2a by putting 4 cubes together, so the potential at the center of cube of side 2a will be
    V_center of cube 2a = 8 V_corner
     
  2. jcsd
  3. Dec 7, 2011 #2
    Make 7 copies of the charged cube of length a. We want to know the potential at the corner of the smaller cube, call that Vc. Arrange the 8 cubes into a larger cube of length 2a. We know that the potential at the center of the larger cube will be 8*Vc. We must now relate that to the known potential at the center of the smaller cube which is Vo somehow using dimensional analysis.

    I'm stuck for now %^(
     
  4. Dec 7, 2011 #3
    i used Mathematica to calculate the potential at the center of the cube,
    when a->2a, the potential will be 4 times larger

    But i can't see it intuitively!
     
  5. Dec 7, 2011 #4
    I'll throw this out and see if it makes sense.

    Suppose we have cubes of uniform charge rho and of various size a. Let the potential at the center of such cubes be given by,

    V = V(Q,R) where Q is the total charge on the cube and R is the length of the cubes edge.

    Dimensionally

    V = constant*Q/R

    If this is true then if we double the edge length then Q --> 8Q and R --> 2*R so V --> 4V

    Am I getting closer?
     
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