Two balls each of mass 0.1 milligram carry identical charges and suspended by two non conducting threads each of length 0.4 metre. In equilibrium each thread makes an angle of 30 degree with the vertical. Find the charge on each ball
Answers
The charge on each ball is q^2 = 2.04 x 10^-17 C
Explanation:
We are given:
Mass of balls = 0.1 milligram
Length of threads = 0.4 m
Angle "θ" = 30°
Solution:
From equilibrium condition.
Tcos(30°) = mg
Tsin(θ) = F(E)
F(E) = K q^2 / 4.l^2 sin^2(30°) = K.q^2 / 2.l^2
Tanθ = F(E) / mg
θ = 30°
Tan(θ) = K.q^2 / 2 l^2 mg
2 l^2 mg / √ 3 . K = q^2
q^2 = 2 x (0.4)^2 (10^-7)(10) / √3 x 9 x 10^9
q^2 = 2 x 1.6 x 10^-7 / √3 x 9 x 10^9
q^2 = 3.2 x 10^-7 / 15.57 x 10^9
q^2 = 0.205 x 10^-7-9
q^2 = 2.05 x 10^-17 C
Thus the charge on each ball is q^2 = 2.04 x 10^-17 C
A pith ball 'A' of mass 9 X 10-5 kg carries a charge of 5 micro C. What must be the magnitude and sign of the charge on a pith ball 'B' held 2 cm directly above the pith ball 'A' , such that the pith ball 'A' remains stationary ?
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