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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TestHub · Accepted Answer

For small angle θ, equilibrium of forces on one pith ball:

d = 2 L sinθ ≈ 2 L θ => θ ≈ d / (2 L)

Horizontal: T sinθ = k q^2 / d^2

Vertical: T cosθ ≈ T = M g (for small θ)

So: M g · θ = k q^2 / d^2

Substitute θ ≈ d / (2 L):

M g · (d / (2 L)) = k q^2 / d^2

Rearrange:

M g d^3 = 2 k q^2 L

Therefore:

d^3 = (2 k q^2 L) / (M g)

and

d = [ (2 k q^2 L) / (M g) ]^(1/3)

Final answer:

d = ((2 k q^2 L) / (M g))^(1/3)

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