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