Question

As shown in the figure, an unknown charge is located at the origin and an electron is located at x = +0.2 nm. If the net electric force on a small positive test charge is zero at x = -1.5 nm, what is the magnitude and sign of the unknown charge? test charge q unknown charge Q electron x=-1.5 nm x=0 x=+0.2 nm C -0.78e C+0.78e C -0.018 +0.018e +28e

Answer 1

From the figure, the magnitude of force acting on the test charge q due to the unknown charge Q is,

$F_{qQ}=k\frac{qQ}{(1.5\times 10^{-9})^{2}}$

The magnitude of force acting on the test charge q due to the electron of charge e is,

$F_{qe}=k\frac{qe}{(3.5\times 10^{-9})^{2}}$

Given the net force on the test charge q is zero. Therefore,

F9Q + Fge = 0

$F_{qQ}=-F_{qe}$

$k\frac{qQ}{(1.5\times 10^{-9})^{2}}=-k\frac{qe}{(3.5\times 10^{-9})^{2}}$

$Q=\frac{-e\times (1.5\times 10^{-9})^{2}}{(3.5\times 10^{-9})^{2}}=-0.18e$

So the unknown charge is -0.18e.

Third option is the correct answer.