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)

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 (=2.05*10^-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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