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

here,

the charge on first particle , q1 = 4.3 uC = 4.3 * 10^-6 C

r1 = 4.7 cm = 0.047 m

mass of second particle , m2 = 6.7 g = 0.0067 kg

the charge on second particle , q2 = 3.7 uC = 3.7 * 10^-6 C

r2 = 7.7 cm = 0.077 m

let the speed of particle be v

using conservation of energy

K * q1 * q2 /r1 = K * q1 * q2 /r2 + 0.5 * m2 * v^2

9 * 10^9 * 4.3 * 10^-6 * 3.7 * 10^-6 * ( 1/0.047 - 1/0.077) = 0.5 * 0.0067 * v^2

solving for v

v = 18.8 m/s

the speed of 3.7 uC charge is 18.8 m/s