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December 22, 2014

December 22, 2014

Posted by **Linn** on Tuesday, April 21, 2009 at 9:38pm.

What I asked was...

An electron is accelerated through a uniform electric field of magnitude 2.5x10^2 N/C with an initial speed of 1.2x10^6 m/s parallel to the electric field.

a) Calculate the work done on the electron by the field when the electron has travelled 2.5 cm in the field.

b) Calculate the speed of the electron after it has travelled 2.5 cm in the field.

c) If the direction of the electric field is reversed, how far will the electron move into the field before coming to rest?

drwls kindly gave me these answer...

a) e E x

b) Assume the work done (above) equals the increase in kinetic energy

c) It will travel until

E e x = intial kinetic energy

I don't quite understand what drwls means by 'x' in the equation eEx. I got an answer for a using W=qEr and got 1x10^.18 J. According to my book, that's correct. For B I got 1.5x10^6m/s, but my book says that the correct answer should be 1.9x10^6 m/s. Additionally, for C, am I using the same kinetic energy as I did in question b? Would I not just get the same answer as b? I'm just really confused and would appreciate further guidance on this problem.

- Physics - urgent, please -
**Linn**, Tuesday, April 21, 2009 at 9:55pmPlease, I've been trying to solve this problem for hours and I still can't get the correct answer, I would really appreciate some more help with this.

- Physics - urgent - due tomorrow -
**Linn**, Tuesday, April 21, 2009 at 10:09pmplease, I honestly keep trying to look at the problem from different points of view, but nothing is working out; I'm running out of ideas as to how I can approach and still, I'm not getting the right answer or understanding drwls' answers...

- Physics - urgent -
**drwls**, Tuesday, April 21, 2009 at 10:16pm<<I don't quite understand what drwls means by 'x' in the equation eEx. >>

x is the distance the electron travels along the E-field. It is 0.025 m in this case. eE is the force, so eEx is the work done.

- Physics - urgent -
**Linn**, Tuesday, April 21, 2009 at 10:20pmOh okay, thank you, now I get the correct answer for c, but b is still not working out, could you please help me with that?

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