Question

Two charges are placed on the x-axis: one, of 2.50 μC, at the origin and the other, of -3.50 μC, at x = 0.600 m (Fig. P21.60). Find the position on the x-axis where the net force on a small charge +q would be zero.

Short answer

The net force on charge +q would be zero when +-q is at distance 3.27 m at negative x-axis.

Step-by-step solution

Position of Charges

The unknown charge is positive like charge q_1,therefore the position of the charge +q must be left to charge q₁, as shown below.

Net Force

The net force that equals zero could be given by

$\begin{array}{r}{F}_{\text{net}}=F_1-F_2\\ 0=\frac{1}{4\pi {\epsilon }_3}\frac{\left|q_1\right||+q|}{d^2}+\frac{1}{4\pi {\epsilon }_{\text{ה}}}\frac{\left|q_2\right||+q|}{(d+0.600\mathrm{m}{)}^2}\\ \frac{\left|q_1\right|}{d^2}=\frac{\left|q_2\right|}{(d+0.600\mathrm{m}{)}^2}\\ \frac{d}{d+0.600\mathrm{m}}=\sqrt{\left|q_1\right|}\end{array}$

Substitution

Put values to find d:

$\begin{array}{r}\frac{d}{d+\mu 600\mathrm{m}}=\sqrt{\frac{\mid \beta G0}{\mid -3.50}\mid }\\ d=3.27\mathrm{m}\end{array}$

Therefore, the net force on charge +q would be zero when +-q is at distance 3.27 m at negative x-axis