0.500 m rCe 60.0 2.00 AwC 4.00 HC re Figure P23.15 Problems 15 and 30. 16. Two s
ID: 1552964 • Letter: 0
Question
0.500 m rCe 60.0 2.00 AwC 4.00 HC re Figure P23.15 Problems 15 and 30. 16. Two small metallic spheres, each of mass m 0.200 g, are suspended as pendulums by light strings of length L as shown in Figure P23.16. The spheres are given the same electric charge of 7.2 nC, and they come to equilibrium when each string is at an angle of 5.000 with the verti Figure P23.16 cal. How long are the strings? 17. Review. In the Bohr theory of the hydrogen atom, an electron moves in a circular orbit about a proton, where the radius of the orbit is 5.29 XExplanation / Answer
There are three forces acting on each sphere:
1) The tension T of the string
2) The weigt W of the sphere (Y axis direction)
3) The force Fe of electrostatic repulsión (X axis direction)
From the equilibrium equations:
Fx = T sin - Fe = 0 ..............(1)
Fy = T cos - W = 0 ...............(2)
T = string tension
Fe = k*q^2 / X^2 = Coulomb’s force
W = M*g = weight of each sphere
M = 0.2 g = 0.0002 kg = mass of the sphere
= 5° = angle
k = 8.99 x 10^9 N m^2 / C^2 = Coulomb’s constant
q = 7.2 nC = 7.2 x 10^-9 C = electric charge of each sphere
g = 9.81 m/s^2 = acceleration of gravity
X = distance between charges.
From (1)
T = Fe / sin ...............(3)
Replacing in (2)
Fe = W*tan = M*g*tan ..............(4)
From Coulomb’s law
Fe = k*q^2 / X^2
From (4), (5)
M*g*tan = k*q^2 / X^2 ............(6)
The length L of the string is
L = X / (2*sin ) ..........(7)
From (6), (7)
L = (q / 2sin)*SQRT [k / (M*g*tan )]
L = (7.2 x 10^-9 C / 2sin5°)*SQRT[8.99 x 10^9 Nm^2/C^2/(0.002 kg*9.81 m/s^2 *tan 5°)]
L = 0.2989 m 0.3 m
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.