Calculating Work for a Dipole: 90J

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SUMMARY

The discussion centers on calculating the work required to bring a 10.0 μC charge from infinity to a specific position near a dipole consisting of ±6.0 μC charges. The correct answer is established as 90J, derived using the potential of an electric dipole formula: V = kq(1/r+ - 1/r-), where k is 9.0 × 10^9 N∙m²/C². Participants clarify the definitions of r+ and r- as the distances from the 10 μC charge to the positive and negative charges of the dipole, respectively, and emphasize the importance of accurate unit conversions in calculations.

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  • Understanding of electric dipole concepts
  • Familiarity with Coulomb's law and electric potential
  • Proficiency in unit conversion techniques
  • Basic algebra for solving equations
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This discussion is beneficial for physics students, educators, and anyone interested in understanding electric dipoles and the work involved in moving charges within electric fields.

HenryHH
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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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