Question

2 pith balls are suspended by 2 strings of 20 cm length separation between them is 5 cm. when in equilibrium separation is 3 cm. find mass and tension on each ball where both the balls have same charge q.

Answer 1

Arrangement are shown in attachment . Let θ is the angle made by string with vertical . Here tanθ = perpendicular/base = √{20² - 1²}/1 = √{399} ≈ 20
e.g., cosθ = base/perpendicular = 20{approximately}/20 = 1
Hence, cosθ = 1 and sinθ = perpendicular/hypotenuse= 1/20

Now, at equilibrium
Tcosθ = mg { weight } -----(1)
Tsinθ = F { electrostatic force between them } ------(2)

Now, F = kq₁q₂/r²
Here , q₁ = q₂ = 2 × 10⁻⁸ C and r = 3cm = 0.03 m

Now, F = 9 × 10⁹ × 2 × 10⁻⁸ × 2 × 10⁻⁸/(0.03)²
= 9 × 10⁹ × 4 × 10⁻¹⁶/9 × 10⁻⁴
= 4 × 10⁻³ N

Now, Tsinθ = 4 × 10⁻³
or, T × 1/20 = 4 × 10⁻³ [ sinθ = 1/20 }
or, T = 4 × 20 × 10⁻³ = 0.08N
Hence, tension in string is 0.08N

Now, Tcosθ = mg
or, T × 1 = m × 10 [ g = 10 m/s²
or, 0.08 = m × 10 ⇒ m = 0.008 Kg

Hence, mass of each ball is 8g