The total electrostatic force on charge A is [tex]28 \mu N[/tex]
Explanation:
The magnitude of the electrostatic force between two charges is given by Coulomb's law:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where:
[tex]k=8.99\cdot 10^9 Nm^{-2}C^{-2}[/tex] is the Coulomb's constant
[tex]q_1, q_2[/tex] are the two charges
r is the separation between the two charges
Here we have three positively charged particles A,B and C, located at the following positions:
[tex]x_A = 0\\x_B = 10 m\\x_C = 20 m[/tex]
The magnitudes of the three charges are:
[tex]q_A = q_B = q_C = 0.5 \mu C = 0.5\cdot 10^{-6}C[/tex]
The force exerted by B on A is to the left (because the force between two positive charges is repulsive), and the force exerted by C on A is also to the left (also repulsive). Therefore, the net force on A is just the sum of the two forces exerted by charges B and C:
[tex]F_A = F_{BA} + F_{CA} = k\frac{q_B q_A}{(x_B-x_A)^2}+k\frac{q_C q_A}{(x_C-x_A)^2}=\\=(8.99\cdot 10^9) \frac{(0.5\cdot 10^{-6})^2}{(10)^2}+(8.99\cdot 10^9) \frac{(0.5\cdot 10^{-6})^2}{(20)^2}=2.8\cdot 10^{-5} N = 28 \mu N[/tex]
Learn more about electric force:
brainly.com/question/8960054
brainly.com/question/4273177
#LearnwithBrainly
