# Homework Help: Voltage and Electro Potential Energy

1. Sep 21, 2006

### himura137

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.

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

2. Sep 21, 2006

### Astronuc

Staff Emeritus
In part a, the potential is measured with respect to some other point, e.g. ground. The particle is at rest at B, so not kinetic energy, then arrives at A or C with some velocity, and a kinetic energy of mv2, and the energy imparted to the particle comes from the electrical field, and the energy is charge (q) * voltage (V) = qV. In the case of A, B, the Voltage difference is VAB = VA-B.

And similar for VCB. Since the particle is + charged, the potential at A and C must be greater than B.

Apply qV = 1/2 mv2.

In part b, one found the increase in potential for each charge with respect to q1. One is correct for q2. However, when q3 is brought to the line, there are q1 and q2 which contribute to the electric field, and when q4 is brought to the line, there are q1, q2 and q3, present, so one must include the effect of each prior charge on the last charge.

3. Sep 21, 2006

### himura137

1) i am confused, i though the particle was at rest at A and V and arrive at B with some velocity and how did you know that potential at A and C (equal) is greater than B

2) i change the equation and got thsi
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
(q)(k)/(.36m) + (q)(k)/(.36m x 2) = V
EPE3 = (V)(1.6 micro columb)

EPE of q4
(q)(k)/(.36m) + (q)(k)/(.36m x 2) + (q)(k)/(.36m x 4) = V
EPE4 = (V)(1.6 micro columb)

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

4. Sep 21, 2006

### Astronuc

Staff Emeritus
1) i am confused, i though the particle was at rest at A and V and arrive at B with some velocity and how did you know that potential at A and C (equal) is greater than B

A positive charge moves from the more positive (+) potential to the less positive (more negative) potential. A negative charge is attracted to the more positive potential.

If one has a battery, there is (-) and (+) terminals, but the potential is measured with respect to some reference, e.g. ground. If there is a 5V potential diffence, that could be between 5 V and 0 V, or 10 V and 5 V, or 105 V and 100 V, . . . . It is the difference in potential that is important.

In part a, the particle starts at rest (no KE) and accelerates to some velocity vB while traveling between A and B, so

q(VA-VB) = 1/2 mvB2,

similarly

q(VC-VB) = 1/2 m(2vB)2

If the potential difference was negative (-V) and the charge was negative (-q), then the energy imparted to the negative charge (-q)(-V) = qV.

5. Sep 21, 2006

### himura137

thanks i got it, how did know that q(delta)V = 1/2mv^2, i was tryin to look for a relationship between velocity and potential but could not figure it out

and can you check my work for #2, i stil could not get the correct answer