Two small spheres each have a mass m of 0.100 g and are suspended as pendulums by light insulating strings from a common point, as shown in the figure. The spheres are given the same electric charge, and the two come to equilibrium when each string is at an angle of 0 = 6.00° with the vertical. If each string is L = 1.00 m long, what is the magnitude q of the charge on each sphere? C

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Chapter6: Gauss's Law
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Problem 55P: The electric field 10.0 cm from the surface of a copper ball of radius 5.0 cm is directed toward the...
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these r two separate questions.
Two small spheres each have a mass m of 0.100 g and are
suspended as pendulums by light insulating strings from a
common point, as shown in the figure. The spheres are given
the same electric charge, and the two come to equilibrium
when each string is at an angle of 0 = 6.00° with the vertical.
If each string is L = 1.00 m long, what is the magnitude q of
the charge on each sphere?
C
q =
Transcribed Image Text:Two small spheres each have a mass m of 0.100 g and are suspended as pendulums by light insulating strings from a common point, as shown in the figure. The spheres are given the same electric charge, and the two come to equilibrium when each string is at an angle of 0 = 6.00° with the vertical. If each string is L = 1.00 m long, what is the magnitude q of the charge on each sphere? C q =
An electric field of magnitude 5.00 x 102 N/C exists at all points just outside the surface of a 2.00-cm-diameter steel ball
bearing and points radially outward. Assume the ball bearing is in electrostatic equilibrium.
What is the total charge Q on the ball?
Q =
What is the surface charge density o on the ball?
=
Question Source: Freedman College Physics
Transcribed Image Text:An electric field of magnitude 5.00 x 102 N/C exists at all points just outside the surface of a 2.00-cm-diameter steel ball bearing and points radially outward. Assume the ball bearing is in electrostatic equilibrium. What is the total charge Q on the ball? Q = What is the surface charge density o on the ball? = Question Source: Freedman College Physics
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