- #1
mitleid
- 56
- 1
We just started applying the concept of potential energy and conservative forces in my electromagnetism class, so I might have a few of these up soon.
Question : Through what potential difference would an electron need to be accelerated for it to achieve a speed of 39.0% of the speed of light, starting from rest? The speed of light is 3.00e8m/s; Mass of electron = 9.11e-31; abs.value(e) = 1.60e-19 C
Ka + Ua = Kb + Ub where K is kinetic energy and U is potential energy of the charge.
Object is at rest, so equation may be simplified to -Kb = Ub - Ua or Kb = -(Ub - Ua).
U = V*q where q is the charge in coulombs and V is the potential. The change in V (from rest to .39 the speed of light) is what I'm looking for...
-[tex]\Delta[/tex]U/q = -(Ub/q - Ua/q) = -[tex]\Delta[/tex]V
I can start substituting now to solve for [tex]P\Delta[/tex]V. I take the above simplified equation based on conservation of energy and divide both sides by the charge q to get
Kb/q = -Ub/q + Ua/q = -[tex]\Delta[/tex]V
Kb = .5(m)v^2 where v = .39(3.00e8).
So my final attempt was .5(9.11e-31(.39(3.00e8)^2))/1.60e-19 = -[tex]\Delta[/tex]V
My answer was around 39000... I'm asked for kV so dividing by 1000 I get 39.0, which seems appropriate, but my answer isn't accepted. The units check out... I have put a negative on the answer, which also seems appropriate according to my above logic, but that often fails me around this time of night!
That abs(e) = 1.6 etc. coulombs seemed a little weird... any idea where I'm going wrong?
Any help is appreciated. :)
Question : Through what potential difference would an electron need to be accelerated for it to achieve a speed of 39.0% of the speed of light, starting from rest? The speed of light is 3.00e8m/s; Mass of electron = 9.11e-31; abs.value(e) = 1.60e-19 C
Ka + Ua = Kb + Ub where K is kinetic energy and U is potential energy of the charge.
Object is at rest, so equation may be simplified to -Kb = Ub - Ua or Kb = -(Ub - Ua).
U = V*q where q is the charge in coulombs and V is the potential. The change in V (from rest to .39 the speed of light) is what I'm looking for...
-[tex]\Delta[/tex]U/q = -(Ub/q - Ua/q) = -[tex]\Delta[/tex]V
I can start substituting now to solve for [tex]P\Delta[/tex]V. I take the above simplified equation based on conservation of energy and divide both sides by the charge q to get
Kb/q = -Ub/q + Ua/q = -[tex]\Delta[/tex]V
Kb = .5(m)v^2 where v = .39(3.00e8).
So my final attempt was .5(9.11e-31(.39(3.00e8)^2))/1.60e-19 = -[tex]\Delta[/tex]V
My answer was around 39000... I'm asked for kV so dividing by 1000 I get 39.0, which seems appropriate, but my answer isn't accepted. The units check out... I have put a negative on the answer, which also seems appropriate according to my above logic, but that often fails me around this time of night!
That abs(e) = 1.6 etc. coulombs seemed a little weird... any idea where I'm going wrong?
Any help is appreciated. :)