The bob of a simple pendulum has mass 2,g and | vedclass

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(D) At equilibrium,the forces acting on the bob are the gravitational force $mg$ acting downwards,the electric force $F_e = qE$ acting horizontally,an

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$	an^{-1}(0.5)$

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(D) At equilibrium,the forces acting on the bob are the gravitational force $mg$ acting downwards,the electric force $F_e = qE$ acting horizontally,and the tension $T$ in the string.For equilibrium,the net force in both horizontal and vertical directions must be zero.$T \sin 	heta = qE$$T \cos 	heta = mg$Dividing the two equations,we get:$	an 	heta = \frac{qE}{mg}$Given: $m = 2\,g = 2 	imes 10^{-3}\,kg$,$q = 5.0\,\mu C = 5 	imes 10^{-6}\,C$,$E = 2000\,V/m$,$g = 10\,m/s^2$.Substituting the values:$	an 	heta = \frac{5 	imes 10^{-6} 	imes 2000}{2 	imes 10^{-3} 	imes 10}$$	an 	heta = \frac{10 	imes 10^{-3}}{20 	imes 10^{-3}} = \frac{10}{20} = 0.5$Therefore,$	heta = 	an^{-1}(0.5)$.

The bob of a simple pendulum has mass $2\,g$ and a charge of $5.0\,\mu C$. It is at rest in a uniform horizontal electric field of intensity $2000\,V/m$. At equilibrium,the angle that the pendulum makes with the vertical is (take $g = 10\,m/s^2$)

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