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A particle, having a charge +5 μC, is initially at rest at the point x = 30 cm on the x axis. The particle begins to move due to the presence of a charge Q that is kept fixed at the origin. Find the kinetic energy of the particle at the instant it has moved 15 cm from its initial position if(a) Q =+15μC and (b) Q = -15μC |
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Answer» From energy conservation, Ui + Ki = Uf + Kf kQq/ri + 0 = kQq/rf + Kf Kf = kQq (1/ri - 1/rf) (a) When Q is +15 μC, q will move 15 cm away from it. Hence rf = 45 cm Kf = 9x 109 x 15 x 10-6 x 5 x 10-6 [ 1/(30 x 10-2) – 1/(45 x 10-2)] = 0.75 J (b) When Q is -15 μC, q will move 15 cm towards it. Hence rf = 15 cm Kf = 9 x 109 x (-15 x 10-6) x 5 x 10-6 [ 1/(30 x 10-2) – 1/(15 x 10-2)] = 2.25 J |
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