Question

A charge of -3.37 μC is fixed in place. From a horizontal distance of 0.0408 m, a particle of mass 9.27 x 10^-3 kg and charge -8.49 μC is fired with an initial speed of 83.0 m/s directly toward the fixed charge. How far does the particle travel before its speed is zero?

Answer 1

According to the given problem,

Using the conservation of linear momentum,

KE_i = mv²/2 = 9.27*10^-3*83²/2 = 31.93 J

PE_i = kq1q2/ri = 8.99*10⁹*-3.37*10^-6*-8.49*10^-6/0.0408 = +6.304 J

Now, using the relation,

P.E_i + K.E_i = P.E_f + K.E_f

P.E_f = 31.93+6.304 = 38.23 J
r_f = kq1q2/P.E_f

= 8.99*10⁹*-3.37*10^-6*-8.49*10^-6/38.23 = 6.73*10^-3 m

The distance traveled = r_f - r_i = 0.0408-0.00673 = 0.0341 m.

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