Minimum Potential?

Question

A charge of 0.883 nC is placed at the origin. Another charge of 0.347 nC is placed at x1 = 13.1 cm on the x-axis.

At which point on the x-axis does this potential have a minimum? Hint:
A necessary condition for the potential to have a minimum is that its derivative is 0

Answer 1

The potential will be minimum at orgin as well as at x=13.1 cm. Obvious.

Answer 2

The potential field around a charge is given by -q/(4*pi*e0*r). The potential is -inf at the location of the charge. Based on your question, (specifying a derivative of zero) that is not the type of minimum you are asked for. However, with both charges the same sign, there will not be any local extrema in the combintation. Are you sure one of the charges isn't negative? If they are both positive, the potential goes to zero as x->infinity, so that's where the min potential lies.

The form of the sum of potentials is a/x + b/(x-x1). The derivative of this is -a.x^2 - b/(x-x1)^2. Setting to 0 gives a(x-x1)^2 + bx^2 =0
(a+b)x^2 - 2a*x1*x + a*x1^2 = 0 This is a quadratic. a=q1, b=q2 If the charge at x1 is negative -.347 then plugging in the numbers and solving using the quadratic formula gives two local extrema, x=35.1cm and x=8.05cm The point at x=35.1 is a local min, the point at x=8.05 is a local max

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