JEE PYQ: Electrostatics Question 10
Question 10 - 2021 (25 Feb Shift 2)
Two small spheres each of mass 10 mg are suspended from a point by threads 0.5 m long. They are equally charged and repel each other to a distance of 0.20 m. Then charge on each of the sphere is $\frac{a}{21} \times 10^{-8}$ C. The value of ‘a’ will be:
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Answer: 20
Solution
At equilibrium, $T\sin\theta = F_e$ and $T\cos\theta = mg$. $\tan\theta = \frac{F_e}{mg} = \frac{kq^2}{r^2 \times mg}$. With $r = 0.2$ m, $\sin\theta = \frac{0.1}{0.5} = \frac{1}{5}$, $\tan\theta \approx 0.2$. Solving: $q = \frac{20}{21} \times 10^{-8}$ C, so $a = 20$.
BETA
