Answer:
0.247 μC
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:
[tex]F_y: T_y - W = 0\\T_y = m*g = 0.002 kg *9.81m/s^2 = 0.01962 N[/tex]
[tex]T_y = T_*cos(50)\\T = \frac{T_y}{cos(50)} = 0.0305 N[/tex]
[tex]T_x = T*sin(50) = 0.0234 N[/tex]
The electric force is given by the expression:
[tex]F = k*\frac{q_1*q_2}{r^2}[/tex]
In equilibrium, the distance between the spheres will be equal to 2 times the length of the string times sin(50):
[tex]r = 2*L*sin(50) = 2 * 0.1m * sin(50) 0.1532 m[/tex]
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.
[tex]F_x = T_x - F_e = 0\\T_x = F_e = k*\frac{q^2}{r^2}[/tex]
[tex]q = \sqrt{T_x *\frac{r^2}{k}} = \sqrt{0.0234 N * \frac{(0.1532m)^2}{9*10^9 N*m^2/C^2} } = 2.4704 * 10^-7 C[/tex]
O 0.247 μC
