Respuesta :
Answer:
Q₁ = Q₂ = 8.84 x 10⁻⁹ C
Explanation:
given,
mass of ball, m = 0.16 g = 1.6 x 10⁻⁴ Kg
ball each other, r = 6.8 cm
Weight of the ball
F_w = m g
F_w = 1.6 x 10⁻⁴ x 9.8
F_w = 1.56 x 10⁻³ N
The tension in each string is a force directed along the length of the string and is the hypotenuse of a right triangle.
we have to find the horizontal component of the forces.
The length of the string,L is 35 cm so, it will be the hypotenuse.
θ be the angle made with imaginary vertical line and the string.
now,
[tex]sin \theta = \dfrac{r\2}{L}[/tex]
[tex]sin \theta = \dfrac{3.4}{35}[/tex]
θ = 5.57°
horizontal component of the force = ?
vertical component of force,F_v = 1.56 x 10⁻³ N
[tex]tan\theta = \dfrac{F_H}{F_v}[/tex]
[tex]tan(5.57^0) = \dfrac{F_H}{1.56\times 10^{-3}}[/tex]
F_h = 1.52 x 10⁻⁴ N
now, each ball will be repelled by
F = 1.52 x 10⁻⁴ N
now calculation of charges
[tex]F = \dfrac{kQ_1Q_2}{r^2}[/tex]
Q₁ = Q₂ because both charge are same
[tex]1.52\times 10^{-4} = \dfrac{9\times10^9Q^2}{0.068^2}[/tex]
Q² = 7.809 x 10⁻¹⁷
Q = 8.84 x 10⁻⁹ C
hence the change on the balls were Q₁ = Q₂ = 8.84 x 10⁻⁹ C
