Question

Please help!!! I have a physics exam tomorrow and I just can't wrap my head around this one!

Answer 1

The total electrostatic force on charge A is $28 \mu N$

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:

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

$q_1, q_2$ are the two charges

r is the separation between the two charges

Here we have three positively charged particles A,B and C, located at the following positions:

$x_A = 0\\x_B = 10 m\\x_C = 20 m$

The magnitudes of the three charges are:

$q_A = q_B = q_C = 0.5 \mu C = 0.5\cdot 10^{-6}C$

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:

$F_A = F_{BA} + F_{CA} = k\frac{q_B q_A}{(x_B-x_A)^2}+k\frac{q_C q_A}{(x_C-x_A)^2}=\\=(8.99\cdot 10^9) \frac{(0.5\cdot 10^{-6})^2}{(10)^2}+(8.99\cdot 10^9) \frac{(0.5\cdot 10^{-6})^2}{(20)^2}=2.8\cdot 10^{-5} N = 28 \mu N$

Learn more about electric force:

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