I already asked this and was given an answer, but I don't fully understand it and so would like a further explanation.

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.

Please, 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.

please, 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...

<<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.

Oh 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?

i do not understand this quastion

I apologize for the confusion. Let's go step by step through the problem and address your questions along the way.

a) To calculate the work done on the electron by the electric field, you correctly used the formula W = qEr, where W is the work done, q is the charge of the electron, E is the electric field strength, and r is the distance traveled in the field.

To clarify what drwls meant by 'x,' it represents the distance traveled. In this case, it is 2.5 cm or 0.025 m. So, the equation becomes W = qE(0.025).

The charge of an electron, q, is given as -1.6x10^-19 C (since electrons have a negative charge). The electric field strength, E, is given as 2.5x10^2 N/C. Plugging in these values gives W = (-1.6x10^-19 C)(2.5x10^2 N/C)(0.025 m).

Solving this equation, you should obtain W ≈ -1x10^-18 J. Note that the negative sign indicates that the work done by the field is negative, indicating that the field is doing work on the electron.

b) To calculate the speed of the electron after traveling 2.5 cm in the field, we can use the work-energy theorem. According to the theorem, the work done on an object is equal to the change in its kinetic energy. So, the work done (which you calculated in part a) should equal the increase in the electron's kinetic energy.

Mathematically, we can write this as W = ΔKE, where W is the work done, and ΔKE is the change in kinetic energy.

Since the work done is -1x10^-18 J (as calculated in part a), and it equals the change in kinetic energy, we can write -1x10^-18 J = ΔKE.

To find the change in kinetic energy, we need to know the initial kinetic energy of the electron. The initial speed of the electron is given as 1.2x10^6 m/s. The kinetic energy of an object can be calculated using the equation KE = (1/2)mv^2, where m is the mass of the object and v is its velocity.

The mass of an electron can be approximated as 9.1x10^-31 kg. Plugging in these values, the initial kinetic energy is KE = (1/2)(9.1x10^-31 kg)(1.2x10^6 m/s)^2.

Solving this equation, you should obtain KE ≈ 7.3x10^-19 J.

Now, we can solve for the change in kinetic energy. ΔKE = -1x10^-18 J - 7.3x10^-19 J. Simplifying this, you get ΔKE ≈ -1.7x10^-18 J.

Since the electron starts with an initial kinetic energy of 7.3x10^-19 J, the change in kinetic energy is negative, indicating a decrease in kinetic energy. Therefore, the final kinetic energy is 7.3x10^-19 J - 1.7x10^-18 J = -1x10^-18 J.

To calculate the final speed, you can use the equation KE = (1/2)mv^2 and solve for v. Rearranging the equation, you have v = √(2KE/m). Plugging in the values, v = √[(2)(-1x10^-18 J)/(9.1x10^-31 kg)]. Calculating this, you should obtain v ≈ 1.9x10^6 m/s.

c) Now, let's address your question about part c. When the direction of the electric field is reversed, the work done on the electron will also change.

Since the work done (W) equals the change in kinetic energy (ΔKE), we can write -W = ΔKE.

In this case, the work done will be positive (since it's opposite to the previous case) and equal to the initial kinetic energy.

So, we can write W = KE (initial).

Using the same initial kinetic energy value of 7.3x10^-19 J (as calculated in part b), we can solve for the distance (x) traveled into the field.

The equation becomes Eex = KE (initial).

Plugging in the values, you have (2.5x10^2 N/C)(x) = 7.3x10^-19 J.

Solving for x, you should obtain x ≈ 2.9x10^-17 m.

Therefore, when the direction of the electric field is reversed, the electron will move approximately 2.9x10^-17 m into the field before coming to rest.

I hope this helps clarify the problem. Let me know if you have any further questions!