Two pith balls of equal mass M and equal charge q are suspended from the same point on long massless threads of length L as shown in figure. If K is the Coulomb's law proportionality constant, then for small values of θ, the distance d between the charged pith balls at equilibrium is

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For small angle θ , equilibrium of forces on one pith ball: d = 2 L sin θ ≈ 2 L θ ⟹ θ ≈ 2 L d ​ Horizontal: T sin θ = d 2 k q 2 ​ Vertical: T cos θ ≈ T = M g (for small θ ) So: M g ⋅ θ = d 2 k q 2 ​ Substitute θ ≈ 2 L d ​ : M g ⋅ ( 2 L d ​ ) = d 2 k q 2 ​ Rearrange: M g d 3 = 2 k q 2 L Therefore: d 3 = M g 2 k q 2 L ​ and d = ( M g 2 k q 2 L ​ ) 1/3 Final answer: d = ( M g 2 k q 2 L ​ ) 1/3

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