Respuesta :
Answer:
Proton A has the larger acceleration
Explanation: According to Coulomb's law, the force of attraction/repulsion (repulsion in this case since the protons and the plane are all positively charged), between two charges/charged bodies, is directly proportional to the product of the two charges and inversely proportional to the square of their distances apart.
Mathematically, F = (Kq1q2)/(r^2) where K = (9 × (10^9))Nm2/C2
For this question, the force of repulsion is the one that drives the motion of the protons away from the plane of infinite positive charge.
Since the two charges (A & B) are both protons, it means they have exactly the same amount of charges and the same amount of mass.
Force of repulsion = Force causing motion
(Kq1q2)/(r^2) = ma.
For the two cases (proton A and B), K = K, charge of proton A, q1A =charge of proton B, q1B. The plane of infinite positive charge is the same for both cases i.e. q2 = q2.
But distance between proton B and the plane of infinite positive charge, rb, = 2 × distance between proton A and the plane of infinite positive charge, ra.
This means the repulsion force experienced by proton A would be more than that experienced by proton B. Fa > Fb. (This is because the bigger rb makes the Fb lesser than Fa)
And since the masses are of equal value, the acceleration experienced by proton A would be higher.
Fa > Fb; acceleration of A > acceleration of B (since mass of A = mass of B)
Solved!
