In a vacuum, a proton moves along a uniform electric field line. At point A of the field with a potential of 445V

In a vacuum, a proton moves along a uniform electric field line. At point A of the field with a potential of 445V, it had a velocity of 10 ^ 5 m / s. Determine the potential of point C of the field, in which its speed will double.

q = 1.6 * 10 ^ -19 Cl.
m = 1.6 * 10 ^ -27 kg.
φ1 = 445 V.
V1 = 10 ^ 5m / s.
V2 = 2 * V1.
φ2 -?
Let us use the law of conservation of energy: q * (φ1 – φ2) = m * (V22 – V12) / 2. Since by the condition of the problem 2 = 2 * V1, then q * (φ1 – φ2) = m * (4 * V1 ^ 2 – V1 ^ 2) / 2.
q * (φ1 – φ2) = m * 3 * V1 ^ 2/2.
q * φ2 = q * φ1 – m * 3 * V1 ^ 2/2.
φ2 = φ1 – m * 3 * V1 ^ 2/2 * q.
φ2 = 445 V – 1.6 * 10 ^ -27 kg * 3 * (105m / s) 2/2 * 1.6 * 10 ^ -19 C = 295 V.
Answer: the potential of the electric field will be φ2 = 295 V.