The potential at location A is 429 V. A positively charged particle is released there from...
The potential at location A is 429 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 783 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.
Homework Answers
vA = 429 v
vC = 783 v
vB = ?
from A to B
using work energy theorem
work = change in KE
(vA-vB)*q = 0.5*m*vB^2 .......(1)
from C to B
at B speed = 2vB
(vC - vB) *q = 0.5*m*4vB^2........(2)
from 1 & 2
(vA-vB)*4 = vC - vB
4*429 - 4VB = 783 - vB
vb = 311 V <<<<<------------answer
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