Question

A particle with a charge of -1.3 uc and a mass of 2.2 x 10-6 kg is released from rest at point A and accelerates toward point B, arriving there with a speed of 37 m/s. The only force acting on the particle is the electric force. (a) Which point is at the higher potential? point A point B Give your reasoning. Negative charge accelerates from a lower potential to a higher potential. Negative charge accelerates from a higher potential to a lower potential. (b) What is the potential difference VB - VA between A and B?

Answer 1

Consider the key Point B poent A 10 = 34 mg) bo At prudent A, the kinetic energy of the charge ej zero But at present B, it has tome non-zero kinetore energy Keneter energy at A KA KEATO Kenetic energy at B KEB= k me = 22.2x106x37²5 = 1.51mJ The increase in keneteć energy of l.sim] is due to decrease in potential energy - potential energy at point a> the potential energy at B The charge on the moving particle is -1.3,4C (-ve) we know potential = potenteral Energy charge (-ve) & potential at Ac petented at B.. we have change en Kenedic energy Ak=k&B-ken = 1.51 m) As there is no external work done on the byhtem Energy will remain Conserved icon where AUPOK=0 ou change in potential energy a su= Bk = - 1.51m5 » UB-UA = -1.51m] Scanned with CamScanner

potential difference VB-VA UB-VA. -1.51mJ - 1. 34c = 1161.54v V Also VB-VA = 161. Sav VB VA poent B is at highane potential that Negative charge accelerates from lower potential to higher potential clear Also VB-VA= 1161.548 as Scanned with CamScanner