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Grade: 11

                        

two identical pith balls are charged by rubbing against each other.they are suspended from a horizontal rod through two strings of length 20cm each, the separation between the suspension points being 5 cm.in equilibrium, the separation between the balls is 3 cm find the mass of each ball and tension in the strings..the charge on each ball has a magnitude 2×10^-8 cm? plz, can u explain each step..plz....cz I need full step ......

6 months ago

Answers : (1)

Arun
24742 Points
							
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
6 months ago
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