rowkem
Mar9-09, 07:34 PM
1. The problem statement, all variables and given/known data
Two stationary positive point charges, charge 1 of magnitude 3.45 nC and charge 2 of magnitude 1.85 nC, are separated by a distance of 50.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.
What is the speed v(final) of the electron when it is 10.0 cm from charge 1?
2. Relevant equations
Ek= (mv^2)/2
U= (k(q1q2))/r
3. The attempt at a solution
I used the following equation:
Ek(f)+U(f) = Ek(i)+U(i)
(mv(f)^2)/2 + (k(q1q2))/r = (mv(i)^2)/2 + (k(q1q2))/r
(9.1x10^-31)(vf)^2)/2 + (((9x10^9)(3.45nC)(-1.6x10^-19))/0.10m) + (9x10^9)(1.85nC)(-1.6x10^-19))/0.40m) = (((9x10^9)(3.45nC)(-1.6x10^-19))/0.25m) + (9x10^9)(1.85nC)(-1.6x10^-19))/0.25m)
v(f) = 7.53 x 10^6 m/s
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So that's my answer. I lose marks if I submit an incorrect answer. That said, I just wanna double check my answer before submitting it. A simple yes or no will suffice. Thanks in advance.
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That seems unreasonably high. Maybe I'm just ignorant but, guess we'll see.
Two stationary positive point charges, charge 1 of magnitude 3.45 nC and charge 2 of magnitude 1.85 nC, are separated by a distance of 50.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.
What is the speed v(final) of the electron when it is 10.0 cm from charge 1?
2. Relevant equations
Ek= (mv^2)/2
U= (k(q1q2))/r
3. The attempt at a solution
I used the following equation:
Ek(f)+U(f) = Ek(i)+U(i)
(mv(f)^2)/2 + (k(q1q2))/r = (mv(i)^2)/2 + (k(q1q2))/r
(9.1x10^-31)(vf)^2)/2 + (((9x10^9)(3.45nC)(-1.6x10^-19))/0.10m) + (9x10^9)(1.85nC)(-1.6x10^-19))/0.40m) = (((9x10^9)(3.45nC)(-1.6x10^-19))/0.25m) + (9x10^9)(1.85nC)(-1.6x10^-19))/0.25m)
v(f) = 7.53 x 10^6 m/s
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So that's my answer. I lose marks if I submit an incorrect answer. That said, I just wanna double check my answer before submitting it. A simple yes or no will suffice. Thanks in advance.
---------------------------------------------
That seems unreasonably high. Maybe I'm just ignorant but, guess we'll see.