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.
Physics - bobpursley, Sunday, July 17, 2011 at 3:57pm
Wouldn't the direction of the E field be opposite to that of the electron?
where did that last substitution come from?
Physics - Chloe, Sunday, July 17, 2011 at 4:01pm
But that only gives me the force. I need the charge to help me find electric field
Physics - CANTIUS, Sunday, July 31, 2011 at 3:03am
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.