Final answer:
When a proton is released from rest and moves towards another fixed proton, it experiences a force of attraction due to their electrostatic interaction. The initial acceleration of the released proton can be found using Newton's second law and Coulomb's law.
Step-by-step explanation:
When a proton is released from rest and moves towards another fixed proton, it experiences a force of attraction due to their electrostatic interaction. The initial acceleration of the released proton can be found using Newton's second law, which states that F = ma, where F is the force acting on the proton, m is its mass, and a is its acceleration. In this case, the electrostatic force is given by Coulomb's law, which is F = k × (q1 ×q2) / r², where k is the electrostatic constant, q1 and q2 are the charges of the two protons, and r is their distance of separation.
To find the initial acceleration of the released proton, we can equate the electrostatic force to the force of attraction and solve for a. Since the released proton is at rest, its initial velocity is zero. Therefore, the initial force acting on it is the electrostatic force. Plugging in the values given in the question:
mass of proton: 1.67 x 10⁻²⁷kg
distance of separation: 2.00 mm = 2.00 x 10⁻³m
q1 = q2 = 1.6 x 10⁻¹⁹ C (charge of proton)
k (electrostatic constant): 8.99 x 10⁹ N.m²/C²
Using these values, we can calculate the initial acceleration of the proton:
Force = (8.99 x 10⁹N.m^2/C²) ×[(1.6 x 10⁻¹⁹ C) ×(1.6 x 10⁻¹⁹ C)] / (2.00 x 10⁻³ m)²
Force = 2.30 x 10⁻¹⁰ N
Acceleration = Force / mass = (2.30 x 10⁻¹⁰ N) / (1.67 x 10⁻²⁷ kg)
Therefore, the initial acceleration of the proton after it is released is approximately 1.38 x 10¹⁷m/s².