- #1
Jus10
- 12
- 2
Homework Statement
[/B]
A) Calculate the electric potential at point A.
B) If a small charged particle with a mass m=5.0 mg and charge q=7.0 nC is released from rest at point A, what will be its final speed vf
Homework Equations
V=Kq/r
Uelec=qV
Uelec=1/2mv2
The Attempt at a Solution
A) I calculated V for each point.
V=(Kq1/r1)+(Kq2/r2)+(Kq3/r3)
V=[(9x109)(2x10-9)/0.03]+[(9x109)(2x10-9)/0.04]+[(9x109)(2x10-9)/0.05]
V=600+450+360
V=1410
For part A, I converted all my nanocoulombs (nC) to Coulombs (C), as well as my centimeters (cm) to meters (m) before plugging them into my equation.
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B) In order to calculate velocity (v), I need to calculate potential energy (Uelec).
Uelec=qV
Uelec=(7x10-9)(1410)
Uelec=9.87x10-6
Again, I convert the 7.0 nC into Coulombs for the 1st half of part B
Now I solved for velocity (v) after converting mass from mg to kg:
Uelec=1/2mv2
9.87x10-6=1/2(5x10-6)(v2)
v2=1/2(5x10-6)(9.87x10-6)
v2=2.4675x10-11
I then took the square root of both sides
v=5x10-6 m/s
It's asking for an assessment of the problem. "How does the velocity compare with a typical bullet's velocity, vbullet=120 m/s?"
I'm not sure there is even anything there to compare. With the velocity of the particle at v=0.000005 m/s, and the velocity of a typical bullet at v=120 m/s, the particle is substantially slower than the average bullet. So I must be doing something wrong when there is nothing there to compare.