Three 2.50-μC charges are placed on tiny conducting spheres at the ends of 1.00 m-long strings that are connected at 120° angles as shown below. The magnitude, in N, of the force on any one of the charges is

Question

Three 2.50-μC charges are placed on tiny conducting spheres at the ends of 1.00 m-long strings that are connected at 120° angles as shown below. The magnitude, in N, of the force on any one of the charges is ​  ​

Quizplus · Accepted Answer

The force between two charges can be calculated using Coulomb's law: 
$F=\frac{kq_1q_2}{r^2}$ where F is the force, k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between the charges. 
To find the net force on one of the charges, we need to calculate the force due to each of the other charges and then add them vectorially. 
Taking the direction of the force to be away from the other charges, the force due to one of the other charges is: 
$F_1=\frac{kq_1q_2}{r^2}$ where r is the distance between the two charges, which is given by the length of the string: r = 1 m. Substituting the given values, we get: 
$F_1=\frac{(9	imes 10^9 Nm^2/C^2)(2.50	imes 10^{-6} C)(2.50	imes 10^{-6} C)}{(1 m)^2}$
$F_1=562.50 N$
Since there are two other charges, with the same magnitude of charge, each of them will exert a force of 562.50 N on the given charge. These forces are acting at 120° angles to each other, so we can use the law of cosines to find the resultant force: 
$F_r^2=F_1^2+F_2^2-2F_1F_2\cos120^\circ$
$F_r^2=(562.50 N)^2+(562.50 N)^2-2(562.50 N)(562.50 N)\cos120^\circ$
$F_r^2= (562.50 N)^2+ (562.50 N)^2-2(562.50 N)^2(-0.5)$
$F_r^2= 2(562.50 N)^2(1+0.5)$
$F_r=(2.5981)(562.50 N)$
$F_r=1461.42 N$
Therefore, the magnitude of the net force on one of the charges is 1461.42 N, which is closest to answer choice B, 3.25 × 10^-11 N.

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