A pendulum bob of mass 30.7 times 10^{-6} , k | vedclass

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Question

(C) The forces acting on the pendulum bob are the gravitational force $mg$ acting vertically downwards,the electric force $F_e = QE$ acting horizontal

Vedclass · Accepted Answer

$5 	imes 10^{-4} \, N$

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(C) The forces acting on the pendulum bob are the gravitational force $mg$ acting vertically downwards,the electric force $F_e = QE$ acting horizontally,and the tension $T$ in the thread.Since the bob is at rest,the tension $T$ must balance the resultant of the gravitational force and the electric force.$T = \sqrt{(mg)^2 + (QE)^2}$Given: $m = 30.7 	imes 10^{-6} \, kg$,$g = 9.8 \, m/s^2$,$Q = 2 	imes 10^{-8} \, C$,and $E = 20000 \, V/m$.$mg = 30.7 	imes 10^{-6} 	imes 9.8 \approx 3.0086 	imes 10^{-4} \, N$.$QE = 2 	imes 10^{-8} 	imes 20000 = 4 	imes 10^{-4} \, N$.$T = \sqrt{(3.0086 	imes 10^{-4})^2 + (4 	imes 10^{-4})^2} \approx \sqrt{9.05 	imes 10^{-8} + 16 	imes 10^{-8}} \approx \sqrt{25.05 	imes 10^{-8}} \approx 5 	imes 10^{-4} \, N$.

$A$ pendulum bob of mass $30.7 \times 10^{-6} \, kg$ and carrying a charge $2 \times 10^{-8} \, C$ is at rest in a horizontal uniform electric field of $20000 \, V/m$. The tension in the thread of the pendulum is $(g = 9.8 \, m/s^2)$.

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