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demonelite123
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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.
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.