Acceleration of an electron through a uniform electric field

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SUMMARY

An electron is accelerated through a uniform electric field of 250 N/C with an initial speed of 1.2 x 106 m/s. The work done on the electron after traveling 2.5 cm is calculated using the formula W = εqΔd, resulting in 1.0 x 10-18 J. The final speed of the electron, after applying the correct kinetic energy equation, is determined to be 1.9 x 106 m/s, correcting the initial miscalculation. The proper approach involves using the equation 1/2 mvf2 = 1/2 mvi2 + W.

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XJellieBX
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Question:
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.

-----------<---------------
------------<--------------
e- ----------------------->
------------<--------------
-----------<---------------

a) Calculate the work done on the electron by the field when the electron has traveled 2.5cm in the field.
b) Calculate the speed of the electron after it has traveled 2.5 cm in the field.

My attempt:
For a), I used W=\epsilonq\Deltad
and I got 1.0x10^-18 J.

For b), I tried
W=\DeltaE_{K}
=0.5mv_{2}^{2}-0.5mv_{1}^{2}
After rearranging the equation to solve for v_{2},
v_{2}=\sqrt{2W/m}+v_{1}
=2.7x10^6m/s
for m, I plugged in the mass of an e-.

However, the answer is supposed to be 1.9x10^6 m/s.
Can someone please tell me what I'm doing wrong?
 
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XJellieBX said:
For b), I tried
W=\DeltaE_{K}
=0.5mv_{2}^{2}-0.5mv_{1}^{2}
This is OK.
After rearranging the equation to solve for v_{2},
v_{2}=\sqrt{2W/m}+v_{1}
But this is not OK.

Try this:

1/2 m v_f^2 = 1/2 m v_i^2 + W
v_f^2 = (2/m)(1/2 m v_i^2 + W)

...and so on.
 
Thank you =)
 

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