Question

A particle with charge 2 mu C is located on the x-axis at the point -10 cm, and a second particle with charge -9 mu C is placed on the x-axis at 10 cm. What Ls the magnitude of the total electrostatic force on a third particle with charge -7 mu C placed on the x-axis at -2 cm ? The Coulomb constant is 8.9875 Times 109 N m^2/C^2.

Answer 1

The electrostatic force is given by

$F=k_e\frac{|q_1||q_2|}{d^2}$

The charge in +10cm push to the left the charge at -2cm. The distance between them is $d=10cm-(-2cm)=12cm= 0.12m$

The charge in -10cm pull to the left the charge at -2cm. The distance between them is $d=-2cm-(-10cm)=8cm= 0.08m$

Since the to forces has the same direction the net force in the charge at at -2cm is

$F=(8.9875\times 10^9N\cdot m^2/C^2)\frac{(2\times 10^{-6}C)(7\times 10^{-6}C)}{(0.08m)^2}+(8.9875\times 10^9N\cdot m^2/C^2)\frac{(9\times 10^{-6}C)(7\times 10^{-6}C)}{(0.12m)^2}$

$F=58.98N$

Toward left