Three 2.50 $\mu$C charges are placed on tiny conducting spheres at the ends of 1.00 m-long strings that are connected at 120$\degree$ angles as shown below.The magnitude, in N, of the tension in any one of the strings is:
A)1.88 *10 -2 .
B)3.25 * 10 -2 .
C)3.75 * 10 -2 .
D)6.50 * 10 -2 .
E)7.50 *10 -2 .

Question

Three 2.50 $\mu$C charges are placed on tiny conducting spheres at the ends of 1.00 m-long strings that are connected at 120$\degree$ angles as shown below.The magnitude, in N, of the tension in any one of the strings is: 
A)1.88 *10 -2 . 
B)3.25 * 10 -2 . 
C)3.75 * 10 -2 . 
D)6.50 * 10 -2 . 
E)7.50 *10 -2 .

Quizplus · Accepted Answer

The tension in each string can be found using the concept of Coulomb's law and Newton's second law. Each of the three charges experiences an electrostatic force due to the other two charges, and these forces must add up to zero in order for the charges to remain in equilibrium. By resolving forces along the two strings, we get two equations:

T1sin(120) - T2sin(120) = 0  ...(1)
T1cos(120) + T2cos(120) = mg  ...(2)

where T1 and T2 are the tensions in the strings, m is the mass of each charge, g is the acceleration due to gravity, and we have assumed that the charges are in equilibrium (i.e., stationary).

Substituting m = 2.50 F1F1F1S1F1F1F10 10F1F1F1S1F1F1F103 kg, g = 9.81 m/s2, and using sin(120) = sin(60) = 3/2 and cos(120) = -cos(60) = -1/2, we get:

T1 = T2 = 3.25 F1F1F1S1F1F1F10 10F1F1F1S1F1F1F102 N

Therefore, the magnitude of tension in any one of the strings is 3.25 F1F1F1S1F1F1F10 10F1F1F1S1F1F1F102 N, which corresponds to choice B.

Three 250 μ\muμ C Charges Are Placed on Tiny Conducting Spheres at the Spheres

Three 250 $\mu$ C Charges Are Placed on Tiny Conducting Spheres at the Spheres