Question

Two charges q1 = −3.20 nC and q2 = +8.36 nC are at a distance of 1.43 µm from each other. q1 is fixed at its location while q2 is released from rest. (b) The charge q2 has a mass m2 = 7.85 µg. What is its speed when it is 0.340 µm from q1?

Answer 1

using energy conservation

change in electric potential energy of q2 = gain in KE

k q1 q2 ( 1/ r1 - 1/ r2) = 0.5 m v^2

9*10^9* 3.2* 8.36*10^-18* ( 1/ (0.34*10^-6) - 1 / ( 1.43*10^-6) ) = 0.5 * 7.85*10^-9* v^2

v = 11727 m/s

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