Question

Use conservation of energy to find the velocity of a free point charge q1 at 22cm if it started at rest 2cm away from a fixed point charge q2 at the origin. q1 = 25 nC, q2 = 15 nC, and mass of q1 = 5g.

Answer 1

here,

q1 = 25 nC = 25 * 10^-9 C

q2 = 15 nC = 15 * 10^-9 C

the initial seperation , r0 = 2 cm

r0 = 0.02 m

the final seperation , r = 22 cm

r = 0.22 m

q1 = 5 g = 0.005 kg

let the velocity of point charge at 22 cm be v

using conservation of energy

K * q1 * q2 * ( 1/r0 - 1/r) = 0.5 * m * v^2

9 * 10^9 * 25 * 10^-9 * 15 * 10^-9 * ( 1/0.02 - 1/0.22) = 0.5 * 0.005 * v^2

solving for v

v = 0.25 m/s

the velocity of point charge is 0.25 m/s