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densk [106]
3 years ago
7

A proton 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.
Physics
1 answer:
NARA [144]3 years ago
8 0

Answer:

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

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