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 the figure above. If k is the Coulomb's law constant, then for small values of θ, the distance d between the charged pith balls at equilibrium is
A. (2kq²L/Mg)1/3
B. (kq²L/Mg)1/3
C. (2kq²L/Mg)1/2
D. (kq²L/Mg)1/2
E. L/4
(GR9277 #83)
Solution:
In Equilibrium:
T cos θ = Mg
T sin θ = Fcoulomb = kq²/d²
with T = tension
For θ ≪, cos (θ → 0) → 1
T = Mg
sin θ = ½d/L
Mgd/2L = kq²/d²
Mgd³ = 2kq²L
d = (2kq²L/Mg)1/3
Answer: A
In Equilibrium:
T cos θ = Mg
T sin θ = Fcoulomb = kq²/d²
with T = tension
For θ ≪, cos (θ → 0) → 1
T = Mg
sin θ = ½d/L
Mgd/2L = kq²/d²
Mgd³ = 2kq²L
d = (2kq²L/Mg)1/3
Answer: A
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