An electron moving in the + x-direction has an initial speed of 3.7*10^6 m/s at the origin. Its speed is reduced to 1.4*10^5 m/s at x = 2cm. Calculate the potential difference between the origin and that point. Which point is at the higher potential?

My work and calculations:

-ΔU = ΔKE
-eΔV = (1/2)(m)(vf)^2 - (1/2)(m)(vi)^2

-(-1.602*10^-19)(ΔV) = (1/2)(9.11*10^-31kg)((1.4*10^5m/s)^2 - (3.7*10^6m/s)^2)

ΔV = -38.8693V = -39V

There is a higher potential at x = 0.

- Is all of this correct?

Good.Because very easily simplified

Your calculations and approach seem correct, but there is a slight mistake in the final statement.

You correctly calculated the potential difference as ΔV = -38.8693V or approximately -39V. Since the potential difference is negative, it indicates that the electron is moving from a higher potential to a lower potential.

Therefore, the point at x = 0 (the origin) is at a higher potential compared to the point at x = 2cm where the speed of the electron is reduced.

Your calculations are correct, but there is a small mistake in your final conclusion. Since the potential difference ΔV is negative (-39 V in this case), it means the potential at x = 0 is higher than the potential at x = 2 cm. Therefore, the correct statement is that there is a higher potential at x = 0.