Aproton is released from rest inside a region of constant, uniform electric field e1 pointing due north. 32.3 seconds after it is released, the electric field instantaneously changes to a constant, uniform electric field e2 pointing due south. 3.45 seconds after the field changes, the proton has returned to its starting point. what is the ratio of the magnitude of e2 to the magnitude of e1? you may neglect the effects of gravity on the proton.

E₂ / E₁ = 521.64 / 5.95 =87.67

Explanation:

Let d be the distance covered inside electric field . Lt q be the magnitude of charge.

Force under field E₁ = q E₁

acceleration = qE₁/ m

d = 1/2 a t²

d = .5 ( qE₁ / m) x 32.3²

d = 521.64 ( qE₁ / m)

Similarly for return journey,

d = .5 x ( qE₂ / m) x 3.45²

d = 5.95x( qE₂ / m)

521.64 ( qE₁ / m) = 5.95x( qE₂ / m)

E₂ / E₁ = 521.64 / 5.95 =87.67

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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