Speed of an electron in an electric field - trying to find my error.

AI Thread Summary
The discussion focuses on calculating the final speed of an electron released between two positive point charges. The user initially applies the conservation of energy equation but encounters a negative value when solving for speed, indicating an error in their calculations. A key point raised is the incorrect treatment of the electron's negative charge in the potential energy calculations. After correcting this, the user arrives at a final speed of 7.53x10^6 m/s, which raises concerns about its plausibility. The conversation emphasizes the importance of accurately incorporating charge signs in electrostatic potential energy equations.
rowkem
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Homework Statement



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?

Homework Equations



Ek= (mv^2)/2
U= (k(q1q2))/r

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)

(9.1x10^-31)(vf)^2)/2 + 5.6x10^-17 = 3.0 x 10^-17

(9.1x10^-31)(vf)^2)/2 = -2.6 x 10^-17

Then my issue: I end up needing to take the square root of a negative number...yah.

If someone could please point me to the error I've made, it would be appreciated. I'm iffy on my "Ek(f)+U(f) = Ek(i)+U(i)". Did I use the correct equations? Please help, thanks.
 
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You are releasing an electron which carries a negative charge. It looks like you haven't carried that into the calculation of U correctly.
 
That would do it. I end up getting 7.53x10^6 m/s. Seems high - thoughts before I submit?
 
rowkem said:
That would do it. I end up getting 7.53x10^6 m/s. Seems high - thoughts before I submit?

I didn't follow your calculation, but off the cuff with mass at 10-31 a little change in potential should go a long way.
 
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