A particle with a charge of -1.4 μC and a mass of 2.9 x 10-6 kg is released from rest at point A and accelerates toward point B, arriving there with a speed of 22 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.
2 answers
yes
When the particle moved, it lost PE and converted it to KE (conservation of energy) Once it reaches point B we can calculate the KE using .5*mv^2
KE= .5*2.9*10^-6*22^2 = 7.018*10^-4
as we know that the amount of work done is equal to the gain in KE, we know that work is the potential difference times the potential difference hence
qV=KE
therefore substitute KE
-1.4*10^-6V=7.018*10^-4
we find Voltage = 501.285volts (j/c)
KE= .5*2.9*10^-6*22^2 = 7.018*10^-4
as we know that the amount of work done is equal to the gain in KE, we know that work is the potential difference times the potential difference hence
qV=KE
therefore substitute KE
-1.4*10^-6V=7.018*10^-4
we find Voltage = 501.285volts (j/c)
1 answer