Just repeating the numbers on the page: L=2m m1=.08kg q1=6uC q2=10uC C=65 V d=0.
ID: 1471609 • Letter: J
Question
Just repeating the numbers on the page:
L=2m
m1=.08kg
q1=6uC
q2=10uC
C=65 V
d=0.24m
A pith ball is also given positive charge and suspended from the ceiling. A second pith ball is also given a positive charge and placed on a rigid rod which is attached to the ceiling. The first pith ball is able to swing out at an angle as a results of the repulsive electrostatic force. The second pith ball is not able to move because the rod is holding it in place. Both pith balls are placed between two charged parallel plates as shown. Use the values of L, m_1, q_1, q_2, V and d given below to determine the distance between the two pith balls, r, as a result of the electrostatic forces on the first pith ball.Explanation / Answer
As m1 is in equilibrium, net focre acting on it must be zero.
Eelctric field between the plates, E = V/d
Let T is the tension in the string.
Apply, Fnety = 0
T*cos(theta) = m1*g -----(1)
Apply, Fnetx = 0
T*sin(theta) - q1*E - k*q1*q2/r^2 = 0
T*sin(theta) = q1*E + k*q1*q2/r^2 ----(2)
take e1uation(2)/equation(1)
tan(theta) = (q1*E + k*q1*q2/r^2)/(m1*g)
m1*g*tan(theta) = q1*E + k*q1*q2/r^2
m1*g*sin(theta) = q1*V/d + k*q1*q2/r^2
m1*g*r/L = q1*V/d + k*q1*q2/r^2
m1*g*r^3 = q1*V*r^2/d + k*q1*q2
0.08*9.8*r^3 = 6*10^-6*65*r^2/0.24 + 9*10^9*6*10^-6*10*10^-6
0.784*r^3 = 1.625*10^-3*r^2 + 0.54
0.784*r^3 - 0.001625*r^2 - 0.54 = 0
on solving the above equation
we get,
r = 0.8838 m <<<<<<-----------Answer
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