Two insulated balls of mass 4.9 g hang from the same support point by massless i

Question

Two insulated balls of mass 4.9 g hang from the same support point by massless insulating threads of length 1(as shown in the diagram.) A total positive charge of q = 7.75x10^-7C is added to the system. Half this charge is taken up by each ball, distributed uniformly, and the balls spread apart to a new equilibrium position. Assuming that the balls hang essentially vertically before the charge is added. what is the tension in each thread before the charge is added? If theta=27.0 degrees, what is the length, l, of each string?

Explanation / Answer

a)T=mg=0.0049*9.81=0.0481N

b)Let the final tension be T.

TSin(theta)=k(q/2)^2/(2LSin(theta))^2

TSin(27)=9*10^9*(7.75*10^-7/2)^2/(2LSin(27))^2

TL^2=0.00361

TCos(theta)=mg

TCos(27)=0.0481

Solving, L=0.054m=5.4cm

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