Question

A particle q1 with the charge of 4.5x10-⁶C is fixed in the space from a distance of 3.7cm , a particle q2 of mass 6.9g and the charge of -3.10x10-⁶C is fired with the initial velocity of 60m/s toward the fixed charge. what is the velocity when it is 1cm away from the q1

Answer 1

The velocity of particle 2 when it is 1cm away from particle 1 is determined as 62.18 m/s.

Force between the charges

The electrostatic force between the two charges is calculated as follows;

F = (kq1q2)/(r²)

F = (9 x 10⁹ x 4.5 x 10⁻⁶ x 3.1 x 10⁻⁶)/(0.037)²

F = 91.71 N

Acceleration of the second particle

F = ma

a = F/m

a = (91.71 N)/(0.0069 kg)

a = 13,291.2 m/s²

Velocity of particle 2 at 1 cm from particle 1

v² = u² + 2as

v² = 60² + 2(13,291.2)(0.01)

v² = 3865.82

v = 62.18 m/s

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