Verifying an Electron's Final Speed After Releasing from Rest

[rowkem]

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)

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 want to 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.

 

[rowkem]

Took a stab at it. Was correct. Thanks everyone.

 

[nlightner]

I would suggest double-checking your calculations and units to ensure accuracy. It is always important to double-check your work and make sure that your answer is reasonable and in line with expected values. In this case, the speed of an electron is typically in the range of 10^6 m/s, so a speed of 7.53 x 10^6 m/s does seem quite high and may warrant further investigation. Additionally, make sure to include units in your final answer to ensure clarity and accuracy.