A particle of charge +9.6 μC is released from rest at the point x = 61...

A particle of charge +9.6 μC is released from rest at the point x = 61 cm on an x axis. The particle begins to move due to the presence of a charge Q that remains fixed at the origin. What is the kinetic energy of the particle at the instant it has moved 21 cm if (a)Q = +66 μC and (b)Q = –66 μC?

Solutions

Expert Solution

Given that

q = 9.6 μC

Q = + 66 μC and - 66 μC

r1 = 0.61 m, r2 = 0.82 m and 0.4 m

a) change in KE = qV = kqQ ( 1 / r1 - 1/r2)
                        = 9 x 10⁹ * 9.6 x 10^-6 * 66 x 10^-6( 1 / 0.0.61 - 1 / 0.0.82)
                        = 2.39 J
b)change in KE = qV
                        = kqQ ( 1 / r1 - 1/r2)
                        = 9 x 10⁹ * 9.6 x 10^-6 * -66 x10^-6 ( 1 / 0.61 - 1 / 0.4)
                        = 4.907 J

Please rate my answer, good luck...

genius_generous answered 1 year ago

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