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Potential difference between 2 points on dipole axis

  1. Oct 4, 2011 #1
    The dipole moment of a water molecule is 6.2 x 10-30 Cm. Find the potential difference VB - VA between 2 points on the axis of the molecular dipole, where points A and B are 8.2 nm and 5.1 nm respectively from the center. Both points are closer to the positive end.

    I used the superposition principle to calculate the potential at each point. so VB = [itex] \frac{kq}{a - 5.1} - \frac{kq}{a + 5.1} [/itex] where a is the distance from each point charge to the center. Also, VA = [itex] \frac{kq}{a - 8.2} - \frac{kq}{a + 8.2} [/itex]. i know that 2aq = 6.2 x 10-30. but now i am having trouble finding a numerical value. when i subtract VB - VA i get these annoying a2 - 5.12 and a2 - 8.22 in the denominator and i can't seem to get rid of them. my answer is messy and in terms of variables while the book's answer is a number. am i missing a piece of information? any help is greatly appreciated.
     
  2. jcsd
  3. Oct 4, 2011 #2

    ehild

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    The dipole potential is given with the assumption that the distance between the positive and negative charge is very small compared to the distance from the dipole. Try to derive the potential at distance x from the dipole centre and take the limit when "a" tends to zero. The superposition principle works, but your formulae are not correct. Check them. Drawing a picture would be very useful.

    ehild
     
  4. Oct 4, 2011 #3
    ah you're right! since i was on the dipole axis i forgot that the distance between the point where i am evaluating the potential and the center is much much larger than the distance between the actual charges. thanks!
     
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