Two small plastic balls hang from threads of negligible mass. Each ball has a mass of 0.14 g and a charge of magnitude q. The balls are attracted to each other. When in equilibrium, the balls are separated by a distance of 2.05 cm and the threads attached to the balls make an angle of 20 degrees with the vertical.(a) Find the magnitude of the electric force acting on each ball.(b) Find the tension in each of the threads.(c) Find the magnitude of the charge on the balls.

a. 4.99 × 10⁻⁴ N b. 1.46 × 10⁻³ N c. 4.83 nC

Explanation:

(a) Find the magnitude of the electric force acting on each ball.

The tension in the string, T is resolved into horizontal and vertical components Tsin20° and Tcos20° respectively, since the string makes an angle of 20° with the vertical.

Let F = electric force and W = mg be the weight of the ball where m = mass of ball = 0.14 g = 0.14 × 10⁻³ kg and g = acceleration due to gravity = 9.8 m/s²

The electric force acts in the opposite direction to the horizontal component of the tension in each string. Since the masses are in equilibrium,

Tsin20° = F (1)

Also, the weight of the ball acts downwards, opposite to the direction of the vertical component of the tension in the string. Since the masses are in equilibrium,

Tcos20° = W = mg

Tcos20° = mg (2)

Dividing (1) by (2), we have

Tsin20°/Tcos20° = F/mg

tan20° = F/mg

F = mgtan20°

So, F = 0.14 × 10⁻³ kg × 9.8 m/s²× tan20°

F = 0.4994 × 10⁻³ N

F = 4.994 × 10⁻⁴ N

F ≅ 4.99 × 10⁻⁴ N

(b) Find the tension in each of the threads.

Since  Tsin20° = F

T = F/sin20°

=  4.99 × 10⁻⁴ N/sin20°

= 4.99 × 10⁻⁴ N/0.3420

= 14.59 × 10⁻⁴ N

= 1.459 × 10⁻³ N

≅ 1.46 × 10⁻³ N

(c) Find the magnitude of the charge on the balls.

From Coulomb's law, the electric force between the masses of charge q and separated by a distance r = 2.05 cm = 2.05 × 10⁻² m

F = kq²/r² where k = 9.0 × 10⁹ Nm²/C²

q² = Fr²/k

q = √(Fr²/k)

q = [√(F/k)]r

Thus,  

q = [√(F/k)]r

q = [√( 4.99 × 10⁻⁴ N/9.0 × 10⁹ Nm²/C²)]2.05 × 10⁻² m

q = [√(0.5544 × 10⁻¹³ C²/m²)]2.05 × 10⁻² m

q = [√(5.544 × 10⁻¹⁴ C²/m²)]2.05 × 10⁻² m

q = [2.355 × 10⁻⁷ C/m]2.05 × 10⁻² m

q = 4.83 × 10⁻⁹ C

q = 4.83 nC

not really sure but thinking 1.54 mm/us