- #1
PrincessYams
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Homework Statement
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Three point charges are arranged as shown here: http://puu.sh/ltvxF.png
a. Find the electric field at point A. (Give as vector, either as i and j or magnitude and direction.)
b. Find the potential at point A.
c. If an electron is placed at point A and released, what is the magnitude of its final velocity?
2. Homework Equations
a. E = (kq)/ r2, Ey = Esin(theta), Ex = Ecos(theta)
b. V = (kq) / r
c. v = (sqrt(2P/m))
CONSTANTS:
k = 8.99 * 109
me = 9.11 * 10-31 kg
The Attempt at a Solution
a. For this one, I calculated all the electric fields from each point. For the left negative charge I got E = 8.9 * 104, which I calculated to be Ex = 72000 and Ey = 54000. Because the right negative charge has the same charge, the electric field is the same. So, the electric fields are the same everywhere EXCEPT the x-component because that is negative, so they cancel out. Currently, I have a Ey = 72000.
For the positive charge, I got E = -5.6 * 104. (I said it was negative because it is going in the negative-y direction?).
So I added all these and got 138400 upwards. My teacher said this was wrong though, where?
b. I took this piece by piece. For the left negative charge, I got -4500 ((k * -2.5 * 108)/.05). For the right negative charge, I got the same thing (same charge and distance). For the middle positive charge, I got +2250 ((k * 1 * 10-8)/ .04). Adding these together, I got 6750 V. Which, again, my teacher said was wrong. I looked it up and found no other way to do it. Even in my notes, it says to do it this way. Where did I do it wrong?
c. I did not really get where to start. I looked up in some other forums around here (will put a link when i find it) and I kept calculating it to be over the speed of light. After b is fixed, I can probably do this one. So ignore it for now.
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