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Sagot :
ANSWER:
0.02182 μC
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 1.057 g = 0.001057 kg
Length spring (l) = 5 cm = 0.05 m
spring stretch (x) = 1.932 cm = 0.01932 m
spring stretch + beads (x1) = 0.286 cm = 0.00286 m
The first thing to calculate is the value of the spring constant k, just like this:
[tex]\begin{gathered} F=k_sx \\ \\ F=mg \\ \\ mg=k_sx \\ \\ k_s=\frac{mg}{x} \\ \\ \text{ We replacing} \\ \\ k_s=\frac{(0.001057)(9.8)}{(0.01932)} \\ \\ k_s=0.5361\text{ N/m} \end{gathered}[/tex]When the charged beads are attached in equilibrium, therefore:
[tex]\begin{gathered} \frac{k\cdot q^2}{(l+x_1)^2}=k_s\cdot x_1 \\ \\ q^2=\frac{(k_s\cdot x_1)(l+x_1)^2}{k} \\ \\ q=\sqrt{\frac{(k_sx_1)(l+x_1)^2}{k}} \\ \\ \text{ We replacing:} \\ \\ q=\sqrt{\frac{\left(0.5361\cdot0.00286\:\right)\left(0.05+0.00286\:\right)^2}{9\cdot10^9}} \\ \\ q=2.182\cdot10^{-8}C\cdot\frac{1\text{ }\mu C}{1\cdot10^{-6}\text{ }C}=0.02182\text{ }\mu C \end{gathered}[/tex]The charge of the beads is 0.02182 μC
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