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Question
Two identical pith balls are charged by rubbing one against the other. They are suspended from a horizontal rod through two strings of length 20 cm 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 the tension in the strings. The charge on each ball has a magnitude 2.0 × 10−8 C.
Solution
As given, q = 2 × 10-8 and r = 3 cm = 3 × 10-2m
Force between two charged particles
∵ `F = (kq_1q_2)/r^2`
= `(9 xx 10^9 xx 2 xx 10^-8 xx 2 xx 10^-8)/((3 xx 10^-2)^2)`
= 4 × 10-3H
from Fig T sin θ = F ...(i)
T cos θ = mg ...(ii)
Tan θ = `F/(mg)`
Here, Tan θ = `P/B = 1/sqrt399 ≈ 1/20`
and `1/20 = F/(mg)`
m = `(20F)/g`
m = 8 g
T = `F/sin θ`
= `(4 xx 10^-3)/(1/20)`
= 8 × 10-2 N
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