lb 0 e 3.33 Repulsion force. A very thin tube contains at its bottom a small pla
ID: 2267452 • Letter: L
Question
lb 0 e 3.33 Repulsion force. A very thin tube contains at its bottom a small plastic ball charged with a charge Q 0.1 C as shown in Figure 3.33. A second, identical ball with identical charge is inserted from above. Assume there is no friction and the glass does not affect the charges on the balls. If each ball has a mass m-1g and the permittivity of the tube may be assumed to be the same as that of air (free space) calculate: (a) The distance between the balls if the tube is vertical (Figure 3.33a) (b) The distance between the balls if the tube is tilted so it makes an angle 30° with the horizontal (Figure 3.33b).Explanation / Answer
Answer :-a) When tube is in vertical position.
Since charges are identical, hence they will repel each other. The upper ball will have two forces-
one acting upward from the below charge, coulomb's force.
second acting downward due to gravity, gravitational froce.
The upper ball will stop falling down when the above two forces will be equal. So we can write-
Coulomb's electrostatic force(in air medium) = Gravitational force on ball
=> (9 x 109) x (0.1 x 0.1 x 10-12) / d2 = (0.001) x 9.8
=> d2 = (9 x 109) x (0.1 x 0.1 x 10-12) / (0.001) x 9.8
=> d = 95.83 mm
b) When tube is kept at an angle of 300 with horizontal.
In this case the gravitational force magnitude changes to m*g*cos(300). The electrostatic force remains same.We can write-
Coulomb's electrostatic force(in air medium) = Gravitational force on ball
=> (9 x 109) x (0.1 x 0.1 x 10-12) / d2 = (0.001) x 9.8 xcos(300)
=> d2 = (9 x 109) x (0.1 x 0.1 x 10-12) / (0.001) x 9.8 x 0.866
=> d = 102.98 mm
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