## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) The proton should be placed at $x = -2.4~cm$ (b) The net force on the electron would also be zero.
(a) We must place the proton to the left of the $2.0~nC$ charge. Then there will be one position such that the forces on the proton due to the two charges are equal and opposite. Let $x$ be the required distance to the left of the $2.0~nC$ charge. We can find $x$: $\frac{k~(2.0~nC)(q_p)}{x^2} = \frac{k~(4.0~nC)(q_p)}{(x+1.0)^2}$ $(x+1.0)^2 = 2.0~x^2$ $x+1.0 = \sqrt{2.0}~x$ $(\sqrt{2}-1)~x = 1.0$ $x = \frac{1.0}{\sqrt{2}-1}$ $x = 2.4~cm$ The proton should be placed at $~~x = -2.4~cm$ (b) If an electron is placed at the same position, the charges on the electron due to the other two forces would also be equal and opposite. Therefore, the net force on the electron would also be zero.