Two small plastic balls hang from threads of negligible mass?

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Each ball has a mass of 0.49g and a charge of magnitude q. The balls are attracted to each other, and the threads attached to the balls make an angle of 20.0 degrees with the vertical, as shown in the figure.


Part A)

Find the magnitude of the electric force acting on each ball.

F = _____ N

Part B)

Find the tension in each of the threads.

T = ____ N

Part C)

Find the magnitude of the charge on the balls.

q = ____ nC

I calculated part B but I keep getting the wrong answer and I only have one attempt left to answer it. I did .49 *9.81/cos10

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The homework program says I’m within 10% of the correct answer, but I just can’t get it right.

2 Answers

  • Part A)

    First, consider the tension on the thread T

    T cos(20) = m g

    Here m = 0.49 * 10^-3 kg

    T = 1 / cos(20) 0.49 10^-3 9.8 m/s^2 = 5.1 10^-3 N

    The electric force F should balance out the x component of the tension T.

    T sin(20) = Electric Force = 1.7 10^-3 N

    Part B)

    T was already calculated in Part A).

    T = 5.1 * 10^-3 N

    Part C)

    The Coulomb force between the charges is:

    F = k * q^2 / r^2

    Here k is the Coulomb constant (=9 10^9), r is the separation (=2.0510^-2 m), and q is the charge.

    q = (F * r^2 / k)^0.5

    In Part A), the electric force was calculated (=1.7 * 10^-3 N).

    q = (1.7 10^-3 (2.05 10^-2)^2 / (9 10^9))^0.5 = 8.9 * 10^-9 C = 8.9 nC

    Answer: 8.9 nC

  • in no way observed the %, yet its ordinary,…basically make certain the stress aspects of the string. The vertical will equivalent that of the load of the ball & the horizontal will equivalent the repulsive stress between the balls. you will discover that out via making use of columb’s regulation (F = kq^2/r^2), the place r is the area seperating the; & would be found out via trigonometric calculations.

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