A bead with charge q1 = 1.90 µC is fixed in place at the end of a wire that makes an angle of θ = 51.3° with t?
A bead with charge
q1 = 1.90 µC
is fixed in place at the end of a wire that makes an angle of θ = 51.3° with the horizontal. A second bead with mass
m2 = 4.20 g
and a charge of 5.55 µC slides without friction on the wire. What is the distance d at which the force of the Earth's gravity on
m2
is balanced by the electrostatic force between the two beads? Neglect the gravitational interaction between the two beads.
Image: http://www.webassign.net/bauerphys1/21-p-078.gif
q1 = 1.9 x10^-6 C
q2 = 5.5 x10^-6 C
θ = 51.3°
m = 4.2 g (4.2 x10^-3 Kg)
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when the parallel weight force to the wire on m2 is equal to the electrostatic force between the two beads.
Electrostatic force between the two beads:
Fe = k (q1 q2)/d²
Force of the Earth's gravity on m2(only
Fp = mg cos(θ)
then:
Fe = Fp
k (q1 q2)/d² = mg cos(θ)
Now solve for the distance (d):
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d = √ k (q1 q2)/(m g cos(θ) ≈ 2.1 m
check the result!
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