My Notes Ask Your Teacher 6. 03.35 points Previous Answers serpset9 23 p016 nva

Question

My Notes Ask Your Teacher 6. 03.35 points Previous Answers serpset9 23 p016 nva 2/100 Submissions Used Two small metallic spheres, each of mass m 0.192 g, are suspended as pendulums by light strings of length L as shown in the figure below. The spheres are given the same electric charge of 7.3 nC, and they come to equilibrium when each string is at an angle of 5.20° with the vertical. How long are the strings? 18.9e-3 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m Need Help? LM..er. Submit Subrnit Assignment Save Assignment Progress

Explanation / Answer

Let q is the charge on each sphere and d is the distance between the spheres.

from figure, d = 2*L*sin(theta)

as the spheres are in equilibirum, net force acting on them must be zero.

Let T is the tension in the strings.

Apply, net forc acting on a sphere in y-direction = 0

Fnety = 0

T*cos(theta) - m*g = 0

T*cos(theta) = m*g -----(1)

Apply, Fnetx = 0

T*sin(theta) - Fe = 0 (here Fe is the electric repulsive force between the spheres)

T*sin(theta) = Fe --(2)

from equations 1 and 2

T*sin(theta)/(T*cos(theta)) = Fe/(m*g)

tan(theta) = Fe/(m*g)

==> Fe = m*g*tan(theta)

k*q^2/d^2 = m*g*tan(theta)

d^2 = k*q^2/(m*g*tan(theta))

d = q*sqrt(k/(m*g*tan(theta))

2*L*sin(theta) = q*sqrt(k/(m*g*tan(theta))

L = (1/2)*(q/sin(theta)*sqrt(k/(m*g*tan(theta))

= (1/2)*(7.3*10^-9/sin(5.2))*sqrt(9*10^9/(0.192*10^-3*9.8*tan(5.2))

= 0.292 m

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