Homework Help: Physics

Posted by Mary on Tuesday, September 25, 2007 at 12:14am.

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.

Physics - bobpursley, Tuesday, September 25, 2007 at 8:25am
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

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