Two tiny spheres of mass in = 4.00 mg carry charges of equal magnitude, 72.0 nC,
ID: 1315610 • Letter: T
Question
Two tiny spheres of mass in = 4.00 mg carry charges of equal magnitude, 72.0 nC, but opposite sign. They are tied to the same ceiling hook by light strings of length 0.530 m. When a horizontal uniform electric field E that is directed to the left is turned on, the spheres hang at rest with the angle theta between the strings equal to 50.0 degree in the following figure. The one on the left The one on the right Part B What is the magnitude E of the field? Express your answer with the appropriate units. E =Explanation / Answer
let r is the distance between two spheres in the equilibrium.
from figure, sin(theta/2) = (r/2)/L
==> r = 2*L*sin(theta/2)
= 2*0.53*sin(50/2)
= 0.448 m
let T is the tension in the thread.
Electric force acting on each sphere, Fe = k*q^2/r^2 + q*E
grvaitational force, Fg = m*g
in the equilibrium, Fnety = 0
T*cos(theta/2) = m*g
==> T = m*g/cos(25)
= 4*10^-6*9.8/cos(25)
= 0.4325*10^-5 N
Force between two charges, F = k*q^2/r^2
= 9*10^9*(72*10^-9)^2/0.448^2
= 2.32*10^-4 N
in the x-direction,
T*sin(theta) + k*q^2/r^2 = q*E
E = (T*sin(25) + k*q^2/r^2)/q
= (0.4325*10^-4*sin(25) + 2.32*10^-4)/(72*10^-9)
= 347.6 N/c <<<<<<-------------------Answer
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