Two protons, A and B, are next to an infinite plane of positive charge. Proton B is twice as far from the plane as proton A. Which proton has the larger acceleration?

A) Proton B.

B) Proton A.

C) Both have the same acceleration.

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!