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OpenStudy (anonymous):
Need help with Calculus I applied maximum and minimum word problem.
The potential energy E of an electric charge q due to another charge q1 at a distance of r1 is proportional to q1 and inversely proportional to r1. If charge q is placed directly between two charges of 2.00nC and 1.00 nC that are separated by 10.0 mm, find the point at which the total potential energy (the sum due to the other two charges) of q is a minimum.

Answer 1

We place the \(q_1 = 2 nC\) charge at the origin, the \(q_2 = 1nC\) charge at \(x = 10 mm\), and our positive test charge q somewhere in the middle at point x.

The potential energy of the \(q, q_1\) system due to their relative position is \(U_1 = \dfrac{k q q_1}{x}\).


The potential energy of the \(q, q_2\) system due to their relative position is \(U_2 = \dfrac{k q q_2}{10^{-2} - x}\).

Giving total potential energy: \(U_T =kq \left( \dfrac{ q_1}{x} + \dfrac{ q_2}{10^{-2} - x} \right)\).

The derivative wrt x is:


\(U'_T =kq \left( - \dfrac{ q_1}{x^2} + \dfrac{ q_2}{ ( 10^{-2} - x)^2} \right)\)

Set to zero and solve.

Wolfram solves \(x \approx 0.0059, 0.034\), in metres of course so convert back to mm if that helps you see it.


The first results lies within the 10mm separation and a second derivative (or a plot) shld confirm it as a min. One charge is pushing left, the other right, so equilibrium.

The second result should be a max. Along, but outside of the 10mm space between \(q_1 \) ad \(q_2\) should lie an optimised position at which PE is maxed out, both charges are pushing the test charge in same direction, you find 34mil as the point at which that push is strongest.

Check it for yourself just to be sure :)