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Calculating electrical potential difference with a sphere

  1. Mar 9, 2012 #1
    1. The problem statement, all variables and given/known data

    A sphere with radius 2.0 mm carries a 1.0 μC charge. What is the potential difference, VB - VA, between point B 4.0 m from the center of the sphere and point A 9.0 m from the center of the sphere? (The value of k is 9.0 × 10^9 N∙m2/C2.)

    2. Relevant equations

    The formula for electrical potential difference: V = (ke x Q)/r

    3. The attempt at a solution

    My professor said the answer is 1300V, but I'm having a hard time arriving at that answer. My understanding is that I should use the formula V = (ke x Q)/r first for point B and then for point A, and then I should subtract the answer I got for point A from the answer I got for point B. However, I'm not sure what to plug in for the "r" value of the formula. After converting from mm to m, the radius of the sphere is .002m. If point B is 4m from the center of the sphere and point A is 9m from the center of the sphere, do I subtract .002 from each of those numbers? Unless I have just incorrectly converted units, I am not getting an answer anywhere near 1300V. Is there some other formula I need to use?
     
  2. jcsd
  3. Mar 10, 2012 #2

    ehild

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    The charged sphere can be taken as if its whole charge were concentrated in the centre. So you have to plug in the given distances from the centre, 4 m and 9 m.

    ehild
     
  4. Mar 10, 2012 #3
    Thanks. So why was the radius of the sphere given? Just to throw off the fact that the problem is actually easier and more straightforward than it looks?
     
  5. Mar 10, 2012 #4

    ehild

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    It is kind of knowledge that the radius of the sphere is irrelevant when you need to find the electric field outside the sphere.

    The charged sphere can be substituted with a point charge in the centre when the charge distribution has spherical symmetry. Not otherwise.

    Sometimes a problem asks the electric field or potential inside a charged sphere. In that case you have to take the charge distribution and the radius into account.

    ehild
     
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