Two identical tennis balls each having mass 'm' and charge 'q' are suspended from a fixed point by threads of length 'l'. What is the equilibrium separation when each thread makes a small angle 'θ' with the vertical?
Two identical tennis balls each having mass 'm' and charge 'q' are suspended from a fixed point by threads of length 'l'. What is the equilibrium separation when each thread makes a small angle 'θ' with the vertical?
Option 1 -
x = ( (q²l)/(2πε₀mg) )¹/³
Option 2 -
x = ( (q²l²)/(2πε₀mg) )¹/³
Option 3 -
x = ( (q²l²)/(2πε₀m²g²) )¹/³
Option 4 -
x = ( (q²l)/(2πε₀m²g) )¹/³
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1 Answer
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Correct Option - 1
Detailed Solution:T sin θ = (1/4πε? ) * q²/ (2lsinθ)²
T cos θ = mg
∴ tan θ = q² / (4πε? mg * 4l²sin²θ)
[tan θ ≈ θ, for small angle]
So, θ³ = q² / (16πε? mgl²)
θ = ( q² / (16πε? mgl²) )¹/³
Also separation = 2l sin θ ≈ 2lθ
= 2l ( q² / (16πε? mgl²) )¹/³
= ( 8q²l³ / (16πε? mgl²) )¹/³
= ( q²l / (2πε? mg) )¹/³
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