Physics please clarify
posted by Mary .
The potential at location A is 379 V. A positively charged particle is released there from rest and arrives at location B with a speed vB. The potential at location C is 789 V, and when released from rest from this spot, the particle arrives at B with twice the speed it previously had, or 2vB. Find the potential at B.
Physics please clarify -
Since the particle accelerates in going from A to B, location B has a lower potential than A. When it goes from C to B, it accelerates again, so B has a lower potential than C.
The potential difference VC - VB is 4 times the potential difference VA - VC, since the particle aquires four times more kinetic energy (twice the velocity) going from C to B than it does going from A to B.
379 - VB = (1/4) (789 - VB)
3/4 VB = 181.75
VB = 242.3 Volts