Question

A particle with charge −5 µC is located on

Answer 1

Answer:

36.25 N

Explanation:

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=9\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the magnitude of the two charges

r is the separation between the two charges

Moreover:

- The force is repulsive if the two  charges have same sign

- The force is attractive if the two charges have opposite sign

In this problem, we have 3 charges:

$q_1=-5\mu C = -5\cdot 10^{-6}C$ is the charge located at $x=+10 cm = +0.10 m$

$q_2=+6\mu C=+6\cdot 10^{-6}C$ is the charge located at $x=-8 cm =-0.08 m$

$q_3=+2\mu C=+2\cdot 10^{-6}C$ is the charge located at $x=-2 cm=-0.02 m$

The force between charge 1 and charge 3 is:

$F_{13}=\frac{kq_1 q_3}{(x_1-x_3)^2}=\frac{(9\cdot 10^9)(5\cdot 10^{-6})(2\cdot 10^{-6})}{(0.10-(-0.02))^2}=6.25 N$

And since the two charges have opposite sign, the force is attractive, so the force on charge 3 is to the right (towards charge 1).

The force between charge 2 and charge 3 is:

$F_{23}=\frac{kq_2 q_3}{(x_2-x_3)^2}=\frac{(9\cdot 10^9)(6\cdot 10^{-6})(2\cdot 10^{-6})}{(-0.08-(-0.02))^2}=30.0 N$

And since the two charges have same sign, the force is repulsive, so the force on charge 3 is to the right (away from charge 2).

So the two forces on charge 3 have same direction (to the right), so the net force is the sum of the two forces:

$F=F_{13}+F_{23}=6.25+30.0=36.25 N$