So working it out that way I found.
ΔV = ΔU / q
ΔU = Uq = (6V)(+3C) = 18J
KE = -PE = -Uq
KE = -18J
KE = 1/2*m*v^2
-18J = 1/2 (2kg) (v^2)
v = i sqrt(18)
V = 4.24 m/s , 4.24i m/s
How do I account for the negative (unreal) component?
Oh you're right. So ΔPE = potential energy of the charge (Uq).
Could I say that ΔPE = -ΔKE? So it would be
ΔKE = -1/2 mv^2?
Just curious was my answer to velocity correct? If the sign was wrong how could I have anticipated that when doing it my (long and unnecessary) way?
Now that I think about it doesn't the KE = PE. Or in this case the KE = potential electric energy (U).
So could I have just skipped all that **** and just used the KE = 1/2 mv^2?
Setting U = 1/2 mv^2 and solving for v?
So the "voltage" is the U due to the E(lectric field). Sorry I am trying to get it in terms that I understand. What is the potential force you're talking about?
If I am reading it right I would say the dKE is 0 because once it reaches B it stops moving / has used all its KE?
Potential difference is voltage is it not?
Electrical potential energy is (U) in J no?
Can you please explain how to do the problem as it seems inferred that I did it wrong.
Homework Statement
A 2kg charge of +3C moves through a potential difference of 6V and begins at rest. What is its final velocity?
Homework Equations
m = 2kg
q = +3C
dV = 6V
Vi = 0 m/s
Vf = ?
F = m*a
(Vf)2 = (Vi)2 + 2(a)(d)
dV = (ke)(q)/(r)
1 N/C is equal to 1 V/m
The Attempt at a...
Ah that makes a lot more sense now. So the reason I am finding the unit vector is I need a vector that is pointing in the same direction as the Force caused by these two charges.
Whenever I see r hat from now on that is referring to the unit vector, correct?
Problem worked out perfectly after...
Homework Statement
Charge A is +2C and is located at <4,0,0>
Charge B is +12.5C and is located at <0,-3,0>
What is the the Force(AonB)
What is the magnitude of F(AonB)
Homework Equations
F = (k)(Qa)(Qb)/(r^2) (r^)
For reference the answers are
< -0.8k, 0.6k, 0 > N (F...