The figure shows two particles with charges q 1 = +4 q and q 2 = q , located at

Question

The figure shows two particles with charges q1 = +4q and q2 = q , located at x=0 and x=L, respectively. L=1cm, q=10nC.

(a) At what point along the x-direction can a positive test charge, q0 , be placed so that it is in equilibrium, i.e. the net force on it is zero? You will get two solutions, but only one makes physical sense. Explain why one of the mathematical solutions must be discarded.

(b) Going back to the original configuration of the two charges, at what x-coordinate can a third charge q3 = +2q be placed to minimize the magnitude of the force on this particle?

Explanation / Answer

part A)

as the magnitude of q1 is higher than q2 and the charges are oposite in sign

test charge must be placed at right of q2 to be in equilibrium

Now, let the position of charge is x

k * 4q/(x)^2 = k * q/(x - L)^2

4/(x)^2 = 1/(x - L)^2

putting value of L

4/(x)^2 = 1/(x - 1)^2

solving for x

x = 2 cm

hence , the test charge must be placed at x = 2 cm

part B)

to minimize the force , we can place it at x = 2 cm where the force on this charge will be zero

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