Okay thanks. My WebAssign has a "Practice Another Version" and within it a "Show Solution". I always come to you guys because you let me figure it out, but this time it doesn't seem to be working. I'm just too incompetent on this topic. So anyway, here's what the "Show Solution" says:
"Let the electric potential at location 3 be V3. From conservation of energy we know that
ΔPE + ΔKE = 0
where KE is the kinetic energy of the proton and PE is the electric potential energy. We also know that
ΔPE = qΔV.We combine the above two equations to get
qΔV + ΔKE = 0
or
q(Vf − Vi) = −(KEf − KEi).
Since the protons are released from rest, we have
KEi = 0.
We are given that
v1 = 3v2
and therefore
KE1f = 9KE2f.We now apply the conservation of energy principle to each proton keeping in mind that proton 1 moves from location 1 to location 3 while proton 2 moves from location 2 to location 3.
Proton 1:
q(V3 − V1) = −KE1fProton 2:
q(V3 − V2) = −KE2fDividing one equation by the other we get
q(V3 − V1)
q(V3 − V2)
=
−KE1f
−KE2f
.
Substitute the values
V1 = 231 V, V2 = 115 V, and KE1f = 9KE2f
and solve for V3.
V3 = 101 V
This makes sense, as a positive charge will gain kinetic energy as it moves from a high potential point to a low potential point."
Why are they dividing the entire equations?