A proton is released from rest in a uniform horizontal electric field. It travels 3.25 m for 5 microsecond. Find the acceleration of the proton and the magnitude of the electric field.
Answers
Answer:
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Given:
A proton is released from rest in a uniform horizontal electric field.
It travels 3.25 m for 5 microseconds.
To Find:
The acceleration of the proton and the magnitude of the electric field.
Solution:
The acceleration of the proton is 2.6 X 10¹¹ m/s² and the magnitude of the electric field is 2.71 X 10³ N/C.
Since the proton is under the influence of a uniform electric field, thus it has a constant acceleration. So, we can use the 2nd equation of motion to find the acceleration of the particle.
The 2nd equation is: s = ut + 1/2 at²
Since given that the proton is released from rest,
⇒ u = 0 m/s
Substituting the values,
3.25 = 1/2 a (5 X 10⁻⁶)²
Or 6.5 = a (5 X 10⁻⁶)²
Or a = 2.6 X 10¹¹ m/s²
From Newton's 2nd law we know that F = ma
and also the electric force experienced by a charged particle is qE. Here q is the charge of the particle and E is the magnitude of the electric field.
Equating the two equations,
qE = ma
The mass of a proton = 1.67 X 10⁻²⁷ kg
The charge on a proton = + 1.6 X 10⁻¹⁹
Substituting values,
E = 1.67 X 10⁻²⁷ X 2.6 X 10¹¹ / 1.6 X 10⁻¹⁹
= 2.71 X 10³ N/C