Physics

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.

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asked by Chloe
  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

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

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    posted by Chloe
  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.

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    posted by CANTIUS

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