Finding the accuracy of an approximated potential

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SUMMARY

The discussion focuses on determining the distance r required for the approximated potential of a dipole to be accurate within 1% of the exact potential. The exact potential is given by the formula V = q1/(4πε0r) + q2/(4πε0(r-d)), while the approximate potential is V ≈ qdcosθ/(4πε0r²). The key inequality to solve is Vexact - Vapproximate ≤ 1.0%(Vexact), leading to the expression 1/r - 1/(r-d) - d/r² ≤ 1.0%. The participants suggest simplifying the fractions to a common denominator to facilitate further calculations.

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Homework Statement


Charges +/- q separated by a distance d make a dipole with dipole moment qd. For large r the potential approaches qdcosθ/(4πε0r²). For θ=0, how large must r be so that the asymptotic function is accurate to 1%?

Homework Equations


Exact V= q1/(4πε0r) + q2 /(4πε0r)
Approximate V= qdcosθ/(4πε0r²)

The Attempt at a Solution


I understand that the following statement must be true:
Vexact - Vapproximate ≤ 1.0%(Vexact)

q1/(4πε0r) + q2 /(4πε0(r-d)) - qdcosθ/(4πε0r²) ≤ 1.0 %.

If we cancel all like constants;
1/r - 1/(r-d) - d/r²) ≤ 1.0 %

I do not understand where to go from here, or even if my approach towards this is correct.

Thank you in advance to all those who are reading.
 
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I think they meant for you to have the two charges at +d/2 and -d/2. When you get the two fractions, you should reduce them to a common denominator. Then, you might get an idea what to do next.

Chet
 

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