Question

A positive charge of 4.30 μC is fixed in place. From a distance of 4.70 cm a from the fixed positive charge, a particle of mass 6.70 g and charge +3.70 μC is released from rest. What is the speed of the +3.70 μC when it is 7.70 cm away from the fixed positive charge?

Answer 1

The sum of kinetic energy(K) and potential energy(U) is constant for the system. i.e K + U = constant

Initially, K =0 .
Potential energy between two charged particles (q1 , q2 ) separated by distance r is given by

U = kq1q2/r

q1 = 4.3μC
q2 = 3.7μC
r = 4.70 cm = 0.047 m

U = (9*10^9*4.3*10^-6*3.7*10^-6)/0.047
= 3.046 J

So initially K+U = J

At r = 7.70 cm
U = kq1q2/r

q1 = 4.3μC
q2 = 3.7μC
r = 7.70 cm = 0.077 m

U = (9*10^9*4.3*10^-6*3.7*10^-6)/0.077
= 1.86 J

K = 1/2mv^2

m = 6.70g = 6.7*10^-3 kg

So,

1/2mv^2 + 1.86 = 3.046

0.5*6.7*10^-3*v^2 + 1.86 = 3.046

=> v = 18.8 m/s