1) The potential at location A is 441 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 805 V, and when released from rest from this spot, the particle arrives at B with twice the speed it previously had, or 2 vB. Find the potential at B.(adsbygoogle = window.adsbygoogle || []).push({});

for this, i place the drawing as

A B C

and i am assuming that A + C = B, i am not sure if thats correct or not and i just added the 2 different potential. But then the Vb and 2Vb isnt used and I am not sure where its used unless its telling me that the distance between A and B is half of B and C.

2) Four identical charges (+1.6 μC each) are brought from infinity and fixed to a straight line. Each charge is 0.36 m from the next. Determine the electric potential energy of this group.

i placed the charges as follow

q1 q2 q 3 q4

and used q1 as the starting point so EPE = 0

then i used the equation V = Kq/r and multiple the V by (1.6 micro columb) to get the EPE. I did that for q2 q3 q4 and changing r as i get closer to q4. Then i added all the EPE together but i am missing something.

EPE of q1 = 0

EPE of q2

(q1)(k)/(.36m) = V

EPE2 = (V)(1.6 micro columb)

EPE of q3

(q1)(k)/(.36m x 2) = V

EPE3 = (V)(1.6 micro columb)

EPE of q4

(q1)(k)/(.36m x 4) = V

EPE4 = (V)(1.6 micro columb)

Add EPE (1 + 2 + 3 + 4) = Total EPE of the group

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# Homework Help: Voltage and Electro Potential Energy

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