Question

Two identical tiny spheres of mass m =2g and charge q hang from a non-conducting strings, each of length L = 10cm. At equilibrium, each string makes and angle θ =50 with the vertical. Find the size of the charge on each spere.

Answer 1

0.247 μC

Step-by-step explanation:

As both sphere will be at the same level at wquilibrium, the direction of the electric force will be on the x axis. As you can see in the picture below, the x component of the tension of the string of any of the spheres should be equal to the electric force of repulsion. And its y component will be equal to the weight of one sphere. We can use trigonometry to find the components of the tensions:

`F_y:  T_y - W = 0\\T_y = m*g = 0.002 kg *9.81m/s^2 = 0.01962 N`

`T_y = T_*cos(50)\\T = (T_y)/(cos(50)) = 0.0305 N`

`T_x = T*sin(50) = 0.0234 N`

The electric force is given by the expression:

`F = k*(q_1*q_2)/(r^2)`

In equilibrium, the distance between the spheres will be equal to 2 times the length of the string times sin(50):

`r = 2*L*sin(50) = 2 * 0.1m * sin(50) 0.1532 m`

And k is the coulomb constan equal to 9 *10^9 N*m^2/C^2. q1 y q2 is the charge of each particle, in this case, they are equal.

`F_x = T_x - F_e = 0\\T_x = F_e = k*(q^2)/(r^2)`

`q = \sqrt{T_x *(r^2)/(k)} = \sqrt{0.0234 N * ((0.1532m)^2)/(9*10^9 N*m^2/C^2) } = 2.4704 * 10^-7 C`

O 0.247 μC