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Vinil7 [7]
2 years ago
7

Three point charges are on the x axis: −1 µC

Physics
1 answer:
Nina [5.8K]2 years ago
5 0

Answer:

0.0078 N

Explanation:

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

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

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges

r is the separation between the two charges

The force is attractive if the two  charges have opposite sign, and repulsive if the two charges have same sign.

In this problem, we have:

q_1 = -1 \mu C = -1 \cdot 10^{-6}C located at x_1=-3 m

q_2=+9 \mu C = +9\cdot 10^{-6}C located at x_2=0 m

q_3 = -5 \mu C = -5\cdot 10^{-6} C located at x_3=+3 m

The force between charge 1 and charge 2 is:

F_{12}=k\frac{q_1 q_2}{(x_2-x_1)^2}=(8.99\cdot 10^9)\frac{(1\cdot 10^{-6})(+9\cdot 10^{-6})}{3^2}=0.0090 N

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

The force between charge 1 and charge 3 is:

F_{13}=k\frac{q_1 q_3}{(x_3-x_1)^2}=(8.99\cdot 10^9)\frac{(1\cdot 10^{-6})(5\cdot 10^{-6})}{6^2}=0.0012 N

And since the two charges have same sign, the force is repulsive, so the force on charge 1 is towards the left.

Therefore, the net  force on charge 1 is:

F=F_{12}-F_{13}=0.0090-0.0012 = 0.0078 N

towards the right.

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