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A proton is released from rest inside a region of constant, uniform electric field E 1 pointing due north. 27.3 s after it is released, the electric field instantaneously changes to a constant, uniform electric field E 2 pointing due south. 3.03 s after the field changes, the proton has returned to its starting point. What is the ratio of the magnitude of E 2 to the magnitude of E 1 ? You may neglect the effects of gravity on the proton.

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Final answer:

The ratio of the magnitude of electric field E2 to E1, when a proton returns to its initial point after the direction of a uniform electric field changes, is approximately 0.0123.

Step-by-step explanation:

A proton is initially at rest in an electric field that points due north. After 27.3 seconds, the electric field changes direction and points due south, and after 3.03 seconds the proton returns to its starting point. To find the ratio of the magnitudes of E2 to E1, we must consider the distances covered by the proton under the influence of both fields given that it starts and ends at the same position.

The distance covered under E1 can be calculated using the formula s = 0.5 × a × t^2, where 'a' is the acceleration and 't' is the time. Since the proton moves for 27.3 s under E1, the distance s1 is s1 = 0.5 × a1 × (27.3)^2. The proton then moves in the opposite direction under E2 for 3.03 s, covering the same distance in opposing direction, thus s2 = 0.5 × a2 × (3.03)^2. Since s1 = s2, we can equate them: 0.5 × a1 × (27.3)^2 = 0.5 × a2 × (3.03)^2.

Furthermore, the acceleration of the proton is directly proportional to the electric field (since a = F/m and F = qE), giving us a1 ∝ E1 and a2 ∝ E2. By simplifying the above equation, we find that a1/a2 = (3.03/27.3)^2, and therefore the ratio of the magnitudes of electric fields is also E2/E1 = (3.03/27.3)^2. Calculating this, we get the ratio E2/E1 approximately equal to 0.0123.

User Geekbuntu
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6.8k points
4 votes

Answer:

E2/E1 =99.2

Step-by-step explanation:

time after release of E1 (t) = 27.3 s

time after release of E2 (t') = 3.03 s

acceleration (a) =
(QE1)/(M)

where

  • Q is the protons charge
  • M is the mass

after 27.3 s

velocity (V) = a x t =
(QE1)/(M) x 27.3 =
(27.3QE1)/(M)

distance to turning point (s) = 0.5a
t^(2) = 0.5 x
(QE1)/(M)x
27.3^(2) =
(372.65QE1)/(M)

now for its return back to its starting point

acceleration (a') =
-(QE2)/(M)

total distance S' = distance to turning point + distance from turning point to starting point

S' = S + vt' + 0.5 a'
t'^(2)

S' =
(372.65QE1)/(M) + (
(27.3QE1)/(M) x 3.03) + (0.5 x
-(QE2)/(M)x
3.03^(2))

S' is the distance at the starting point and = 0

0 =
(372.65QE1)/(M) + (82.72QE1)/(M)-(4.59QE2)/(M)


(4.59QE2)/(M)=(372.65QE1)/(M) + (82.72QE1)/(M)

multiplying both side by M/Q we have

4.59.E2 = 372.65E1 + 82.72E1

4.59.E2 = 455.37E1

E2/E1 = 455.37 / 4.59

E2/E1 =99.2

User Benjismith
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7.5k points