In the figure above, the distance between charges A and B and the distance between charges B and C are both 9.50 cm.If A = 1.50 nC, B = -7.50 nC, and C = 3.50 nC, what is the net electric force on A?
Question
In the figure above, the distance between charges A and B and the distance between charges B and C are both 9.50 cm.If A = 1.50 nC, B = -7.50 nC, and C = 3.50 nC, what is the net electric force on A?
Solution
To solve this problem, we need to use Coulomb's Law, which states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The formula for Coulomb's Law is:
F = k * |q1*q2| / r^2
where: F is the magnitude of the force between the charges, k is Coulomb's constant (8.99 * 10^9 N m^2/C^2), q1 and q2 are the amounts of charge, r is the distance between the charges.
First, we calculate the force between A and B:
F_AB = k * |q_A*q_B| / r_AB^2 = (8.99 * 10^9 N m^2/C^2) * |(1.50 * 10^-9 C) * (-7.50 * 10^-9 C)| / (0.095 m)^2 = -0.0000112 N
The negative sign indicates that the force is attractive.
Next, we calculate the force between A and C:
F_AC = k * |q_A*q_C| / r_AC^2 = (8.99 * 10^9 N m^2/C^2) * |(1.50 * 10^-9 C) * (3.50 * 10^-9 C)| / (0.19 m)^2 = 0.00000441 N
The positive sign indicates that the force is repulsive.
Finally, we add the two forces to find the net force on A:
F_A = F_AB + F_AC = -0.0000112 N + 0.00000441 N = -0.00000679 N
So, the net electric force on A is -0.00000679 N, which means it is attracted to B and repelled by C, with a stronger attraction to B.
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