A proton is released from rest inside of constant, uniform electric field E 1 po

Question

A proton is released from rest inside of constant, uniform electric field E1 pointing due North.

What is the ratio of the magnitude of E2 to the magnitude of E1? You may neglect the effects of gravity on the proton

A proton is released from rest inside a region of constant, uniform electric field E1 pointing due North. 17.3 seconds after it is released, the electric field instantaneously changes to a constant, uniform electric field E2 pointing due South. 7.65 seconds after the field changes, the proton has returned to its starting point. What is the ratio of the magnitude of Ez to the magnitude of E? You may neglect the effects of gravity on the proton Number

Explanation / Answer

Mass of the proton m = 1.67 * 10-27 kg

Charge of the proton q = 1.6 * 10-19 C

Let the initial electric field be E1

Now   

F = q * E, and F = ma,

So, q* E = ma

Then a = q * E/m ………….. ( 1 )

Let the initial velocity of the proton u be zero
Let the acceleration be a1

Here a1 = q * E1 /m
So ,
a = (1.6 × 10-19 )(E) / (1.67 × 10²7 )
= (E)(9.58 × 10) m/s².

After 17.3 s,

v = a1 * t

   = E * 9.58 × 10 * 17.3
   = (E)(1.65 × 109) m/s.

Position of the proton after 17.3 sec ,

X0 = ½a * t² ( since proton is at rest u * t = 0 )

= ½ * (E) * (9.58 × 10) * (17.3 )²

= (E) * (1.43 × 1010) m.

So we can conclude that 7.65 s after the field reversal, the proton has traveled (E)1.43 × 1010 m back to the initial point.

So the net displacement x = 0

   = (E)(1.43 × 1010) + (E)(1.65 × 10)(7.65) -         ½ * a* (7.65)².

Where a2 is acceleration after reversal and opposite in direction to initial motion and so negative
a = 2 * [(E) * (1.43 × 10¹0 ) + (E * 1.65 * 109 * 7.65 )] / ( 7.65 )2

     = E1 * 0.09193 * 1010 m/s2
From ( 1 ) , we can write a2 / a1 = E2 / E1

  So , a/a = ( E* 0.09193 * 1010 ) / (E * 9.58 × 10)

                  = 9.6

So the ratio of the electric fields E2 / E1 = 9.6

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