Two identical simple pendulums each of length | vedclass

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Question

(A) Let $L = 5 \, cm = 0.05 \, m$ be the length of the string and $d = 6 \, cm = 0.06 \, m$ be the distance between the bobs. At equilibrium, the bob

Vedclass · Accepted Answer

$0.12$

Vedclass · Answer

(A) Let $L = 5 \, cm = 0.05 \, m$ be the length of the string and $d = 6 \, cm = 0.06 \, m$ be the distance between the bobs. At equilibrium, the bob is acted upon by three forces: tension $T$, gravitational force $mg$, and electrostatic force $F_e = \frac{kq^2}{d^2}$. Let $	heta$ be the angle the string makes with the vertical. From the geometry, $\sin 	heta = \frac{d/2}{L} = \frac{3 \, cm}{5 \, cm} = 0.6$. Thus, $\cos 	heta = \sqrt{1 - \sin^2 	heta} = 0.8$. In equilibrium, $	an 	heta = \frac{F_e}{mg}$. Substituting the values: $	an 	heta = \frac{0.6}{0.8} = 0.75$. $F_e = \frac{(9 	imes 10^9) 	imes (2 	imes 10^{-6})^2}{(0.06)^2} = \frac{9 	imes 10^9 	imes 4 	imes 10^{-12}}{0.0036} = \frac{36 	imes 10^{-3}}{36 	imes 10^{-4}} = 10 \, N$. Now, $mg = \frac{F_e}{	an 	heta} = \frac{10}{0.75} = \frac{10}{3/4} = \frac{40}{3} \, N$. $m = \frac{40}{3 	imes 10} = \frac{4}{3} \, kg \approx 1.33 \, kg$. (Note: Given the options provided in the original prompt were incomplete, the calculated value is $4/3 \, kg$).

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