Question

A proton is released from rest in a uniform electric field. After the proton has traveled a distance of 10.0 cm, its speed is 1.4 x 106 m/s in the positive x direction. Find the magnitude and direction of the electric field.

Answer 1

Wouldn't the direction of the E field be opposite to that of the electron?

Eq=force=mass*acceleration=mass (velocity^2/(2*distance))

where did that last substitution come from?

Vf^2=2ad or
a=Vf^2/2d

Answer 2

But that only gives me the force. I need the charge to help me find electric field

Answer 3

Because the charge on the proton is positive, the direction of the acceleration is the same as the direction of the electric field, the positive x direction.

v^2 = v_0^2 + 2 a x
(one of the equations of motion)
Solving for a, and using v_0 = 0
a = v^2/2x
This is the source of the substitution used in the solution.

Finally, the charge of a proton is the same (in magnitude) as the charge of an electron.