Answer:
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
