A solid metal ball of radius 1.5 cm bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing a
1 answer:
We have that the electric field at the center of the metal ball due only to the charges on the surface of the metal ball is
From the question we are told that
A solid metal ball of radius 1.5 cm
bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing
uniformly distributed charge of -7 nC
The distance between the centers of the balls is 9 cm
Generally the equation for the electric field is mathematically given as
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Answer:
Explanation:
Let the charge on the ball bearing is q.
charge on glass bead, Q = 20 nC = 20 x 10^-9 C
Force between them, F = 0.018 N
Distance between them, d = 1 cm = 0.01 m
By use of Coulomb's law in electrostatics
By substituting the values
Thus, the charge on the ball bearing is
Answer:
a) v₂ = 4.2 m/s
b) v₂ = 5 m/s
Explanation:
a)
We will use the law of conservation of momentum here:
where,
m₁ = m₂ = mass of bowling pin = 1.8 kg
u₁ = speed of first pin before collsion = 5 m/s
u₂ = speed of second pin before collsion = 0 m/s
v₁ = speed of first pin after collsion = 0.8 m/s
v₂ = speed of second after before collsion = ?
Therefore,
<u>v₂ = 4.2 m/s</u>
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b)
We will use the law of conservation of momentum here:
where,
m₁ = m₂ = mass of bowling pin = 1.8 kg
u₁ = speed of first pin before collsion = 5 m/s
u₂ = speed of second pin before collsion = 0 m/s
v₁ = speed of first pin after collsion = 0 m/s
v₂ = speed of second after before collsion = ?
Therefore,
<u>v₂ = 5 m/s</u>