Doubt regarding direction of movement of electron wrt a field [closed]

Question

An electron with an initial speed of $4.0 × 10^ 6 m/s$ is brought to rest by an electric field. The mass and charge of an electron are $9 × 10 ^{-31} kg$ and $1.6 × 10 ^{19}C$, respectively. Identify the correct statement

(A) The electron moves from a region of lower potential to higher potential through a potential difference of 11.4 V.

(B) The electron moves from a region of higher potential to lower potential through a potential difference of 11.4 V.

(C) The electron moves from a region of lower potential to higher potential through a potential difference of 45 V.

(D) The electron moves from a region of higher potential to lower potential through a potential difference of 45 V

My approach: $Q*V = 1/2 * m * v^2$(wor done by electric field is used to bring electron to rest)

So $ V = \dfrac{(1/2 * 9 * 10^{-31} * (4*10^6)^2)}{1.6 * 10^{-19}} = 45$ volts

Now, my doubt is in option C and D. I thought that it would be option C as an electron always moves from a lower to higher potential. But the answer is given as option D.

Answer 1

as an electron always moves from a lower to higher potential

Why?

In fact, this electric field will accelerate an electron in the direction of increasing electric potential and isn't that the case here?

Yes, if an electron is initially at rest, the electron will start moving in the direction of increasing electric potential.

But here, the electron is slowing down in the electric field and so it must be initially moving from high to low potential, correct? Note that after coming to a (instantaneous) stop in this field, the electron begins moving in the direction of higher potential (unless something captures it at the moment it stops).