potential energy question - physics?
The potential at location A is 390 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 799 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.
Update:the answer is 254 V. Just need an equation to get to this point so that i can do more problems like this one.
Answers & Comments
Verified answer
The kinetic energy of the first particle is (1/2)mv^2.
The second particle has twice the speed, so 4 x the kinetic energy. So the potential difference C - B must be 4 times the potential difference A - B.