Question

12. A particle with a charge of 5.5 x 10-8C is fixed at the origin. A particle with a charge of -2.3 x 10 8C is moved from x = 3.5 cm on the x axis to y = 4.3 cm on the y axis. The change in potential energy of the two-particle system is: a) 3.1 x 10-3J b) -3.1 x 10-3 J c) 6.0 x 10 SJ d) -6.0 x 10-5T e) OJ

Answer 1

Ans:- Potential energy is nothing but the energy possessed by a boog by virtue of its position relative to others. The potential energy for two particles is given as U= Kad kq, 92 where K: Roulomb constant, a = charge of first particle 42 = charge of second particle, & distance between a charged partida Initially we will calulate potential energy and later we use concept of change in potential energy to find the change in potential energy of a two -particle system Consider the particles are in x-axis , as per the given data 9. 25.5*108c, 92 = -2.3x108€ , K=9x109 Nm / m² r = 3.5cm, r = 4.3 cm. - The potential energy at x-asis is U = kq,92 (9x109 Nim?/m2) (-13x108) (+5.5x108) 315w) 1152m ) V = -3.25 X10"] The potential energy in tanis is, 3 I consider partite in 4-ares) (Gx10? N. m/on) (5.5x108 ) (-2.3x108) T (4-3) (tom) ---2,65x1043) we have calculated the Calculated the potential potential enerair enerain of of particles in x-ans and 4-axis. Therefore the change in potential energy is AU=Uz-u, = (-2.658154J) - (-3.25x10 "J) DU = 0.6x1045 The change in potentia Du= 6x1657 Answer is "C". energy