A charged particle of mass m=1.00 10^-4 kg and charge Q= +1 nC travels horizonta

Question

A charged particle of mass m=1.00 10^-4 kg and charge Q= +1 nC travels horizontally towards the fixed charged particle of charge 4Q= +4 nC. When the traveling particle is 0.50 meters away, it has speed of 0.10 m/s (see the sketch below). Neglect influence of gravity! What can be said about the nature of the force that 4Q exerts on Q? State two most relevant facts! How close will the moving particle Q come to the fixed particle 4Q? After the moving particle is deflected by 4Q, what will be the greatest speed that it will have? Assume that the moving particle travels horizontally for the whole time!

Explanation / Answer

4Q is resting on stand.

as this is not moving , so it will create any magnetic field.

hence no magnetic force on Q.

4Q exerts electric force on Q.

{ gravitational force also , but it has very very less value compare to electric field }

1. electric force - 4Q will repel Q.

2. at given instance,

KEi = m v^2 / 2   = (10^-4) (0.10^2) /2 = 5 x 10^-7 J

Pei = k(Q)(4Q) / 0.50 = (9 x 10^9 x 4 x 10^-9 x 10^-9 ) / (0.50) = 7.2 x 10^-8 J

at closeset point,

KEf = 0

PEf = 4kQ^2 / d

Applying energy conservation,

5 x 10^-7 + 7.2 x10^-8 = 0 + (4 x 9 x 10^9 x 10^-18 ) / d

d = 0.0629 m   Or 6.29 cm

3. total energy = 5 x 10^-7 + 7.2 x10^-8

finally this will convert all into KE.

5.72 x 10^-7 = (1 x 10^-4) v^2 /2

v = 0.107 m/s

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