Answer to Question #132399 in Mechanics | Relativity for Jessica

Question

A particle with a charge of -2.2 μC and a mass of 3.3 x 10^-6 kg is released from rest at point A and accelerates toward point B, arriving there with a speed of 63 m/s. The only force acting on the particle is the electric force. What is the potential difference VB - VA between A and B? If VB is greater than VA, then give the answer as a positive number. If VB is less than VA, then give the answer as a negative number.

Answer 1

By energy conservation law,

Energy at A = Energy at B

$V_Aq = V_Bq + \frac{1}{2}mv^2$

$V_B - V_A = - \frac{mv^2}{2q} = \frac{3.3*10^{-6}*63*63}{2*2.2*10^{-6}} = 2976.75V$

This means that $V_B > V_A$