Calculating Work for a Dipole: 90J

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To calculate the work required to bring a 10.0 μC charge to a specific position near a dipole, the potential of the dipole must be correctly evaluated using the formula V = kq(1/r+ - 1/r-). The distances r+ and r- refer to the distances from the 10 μC charge to the positive and negative charges of the dipole, respectively. The correct answer for the work done is 90J, but some participants are encountering issues with their calculations, often resulting in values that are orders of magnitude off. This discrepancy may stem from errors in unit conversions or miscalculations in the distances used in the potential formula. Clarifying the calculations and ensuring accurate unit handling is essential for arriving at the correct result.
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Homework Statement



A dipole with ±6.0 μC charges is positioned so that the positive charge is 1.0 mm to the right of the origin and the negative charge is at the origin. How much work does it take to bring a 10.0 μC charge from infinity to a position x = 3.0 mm, y = 0.0 mm? (The value of k is 9.0 × 10^9 N∙m2/C2.)


Homework Equations



The potential of an electric dipole: V = kq(1/r+ - 1/r-)

The Attempt at a Solution



I think I know how to do this problem, but I have some questions. Are r+ and r- the distances the 10 uC charge is from the positive and negative dipole charges, respectively? The answer to the problem is 90J, but I keep getting something like .0000000009, so I'm wondering what I'm doing wrong here.
 
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Usually when one's result is off by some multiple of 10 it's due to a problem with unit conversions. Can you show your math?
 

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