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plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz solve question no 28

YASH AHUJA , 6 Years ago
Grade 11
anser 1 Answers
Eshan

Last Activity: 6 Years ago

568-1847_Untitled.png
From figure, sin\theta=\dfrac{0.14}{0.5}=0.28
From equilibrium, Tsin\theta=F_{el}
Tcos\theta=mg
\implies F_{el}=mg tan\theta \implies \dfrac{1}{4\pi\epsilon_0}\dfrac{q^2}{(0.28)^2}=10^{-5}\times 10\times 0.29
\implies q=1.58\times 10^{-8}C

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