Question

two identical pith balls are changed by rubbing against each other. they are suspended from a horizzontal rod through two strings of length 20cmeach,the sepration between suspension points being 5cm. in equilibrium, the sepresion btween the ball is 3cm. find mass of each ball and tension.charge on each ball is 2*10-8.

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

Answer 2

Answer:

Explanation:

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