Question

A particle, having a charge +5 c, is initially at rest at the point x = 30 cm on the x axis. The particle begins to move due to the presence of a charge q that is kept fixed at the origin. Find the kinetic energy of the particle at the instant it has moved 15 cm from its initial position if (a) q =+15c and (b) q = -15c

Answer 1

kinetic energy = |change in electrostatic potential energy|.

(a) when q = +15c
repulsive force acts between 5c and 25c
so, final distance, r = (0.3 + 0.15) = 0.45
initial Electrostatic potential energy , E = k(15)(5)/(0.30)
final electrostatic potential energy, E' = k(15)(5)/(0.45)
now, change in electrostatic potential energy, ∆E = E' - E = k(15)(5)/0.45 - k(15)(5)/0.3 = -k(75)/0.9 = -9 × 10^9 × 750/9 = -7.5 × 10¹¹ J

so, kinetic energy = 7.5 × 10¹¹ J

(b) when q = -15c
attraction force acts between charges
so, final distance , r = 0.3 - 0.15 = 0.15
initial Electrostatic potential energy , E = k(-15)(5)/(0.30)
final electrostatic potential energy, E' = k(-15)(5)/(0.15)
now, change in electrostatic potential energy, ∆E = E' - E = -k(15)(5)/0.3 = -k(250) = -2.25 × 10¹² J

so, kinetic energy = 2.25 × 10¹² J