Two hard rubber spheres, each of mass m = 15.4 g, are rubbed with fur on a dry d

Question

Two hard rubber spheres, each of mass m = 15.4 g, are rubbed with fur on a dry day and are then suspended with two insulating strings of length L = 4.80 cm whose support points are a distance d = 2.76 cm from each other as shown in the figure below. During the rubbing process, one sphere receives exactly twice the charge of the other. They are observed to hang at equilibrium, each at an angle of = 10.5° with the vertical. Find the amount of charge on each sphere. (Enter your answers from smallest to largest.)

Explanation / Answer

as they have same charges, they will repel each other.

so the left sphere will be moved to the left of vertical and right sphere will move to the right of vertical

so distance between them will increase each by 4.8*sin(10.5)=0.8747 cm

then total distance between the charges=2.76+2*0.8747=4.5094 cm

let charge on left sphere is q and right sphere is 2*q.
so electric force between them=k*q*2*q/(4.5094*0.01)^2=8.85*10^12*q^2

now let the tension in the wire be T.

then its vertical component will balance the weight of the sphere.

and horizontal component will balance the electric force.

hence T*cos(10.5)=15.4*0.001*9.8

T=0.15135 N

and T*sin(10.5)=electric force=8.85*10^12*q^2

q=56.22 nC

higher charge is 2*q=112.44 nC

so two charges are 56.22 nC and 112.44 nC

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