A proton starts from rest, and moves 1 meter through a constant electric field. The potential difference between its starting and ending points is 500 V. How fast is the proton moving at the end? How large is the electric field? How long does it take the proton to complete its journey? [Late clarification: since the field is accelerating the particle from rest, you can assume that the direction of the particle's motion is parallel to the field.]

E = - 500 V / m,  v = 30.95 10⁴ m / s   and    t = 6.46 10⁻⁶ s

Explanation:

For this problem we use the relation

           ΔU = - E s

           E = -ΔU / s

           E = - 500/1

           E = - 500 V / m

Now we can look for the proton approach

           F = q E

let's use Newton's second law

           F = m a

           a = F / m

           a = q E / m

Now let's use kinematics relations, where the proton starts from the rest

v₀ = 0

           v² = v₀² + 2 a x

           v = √2 q E / m x

           v = √ (2 1.6 10⁻¹⁹ 500 / 1.67 10⁻²⁷  1)

           v = √ (958.08 108)

           v = 30.95 10⁴ m / s

for time let's use the equation

          x = v₀ t + ½ to t2

          t = √2x / a

          t = √ (2x m / qE)

          t = √ (2 1 1.67 10⁻²⁷ / (1.6 10⁻¹⁹ 500))

          t = √ (0.004175 10⁻⁸)

          t = 0.0646 10⁻⁴ s

          t = 6.46 10⁻⁶ s

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s