The potential at location A is 370 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 790 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.
Note that the energy from Vcb is four times (2vb)^2 than Vab.
Energy = kVq, since k and q are constants here, then
energyVab= 1/4energyVcb
Vab=1/2Vcb
(370-Vb)= 1/4(790-Vb)
solve for Vb