Question

As shown, two protons are launched with the same speed from point 1 inside a parallel-plate capacitor. One proton moves along the path from 1 to 2 , the other from 1 to 3 . Points 2 and 3 are the same distance from the positive plate.

a. Is \(\Delta U_{1 \rightarrow 2}\), the change in potential energy along the path \(1 \rightarrow 2\), larger than, smaller than, or equal to \(\Delta U_{1 \rightarrow 3}\) ? Explain.

b. Is the proton's speed \(v_{2}\) at point 2 larger than, smaller than, or equal to the proton's speed \(v_{3}\) at point 3 ? Explain.

Answer 1

The electric field lines are directed from the positive plate to the negative plate—we knew that the equipotential surfaces are perpendicular to the electric field lines. Thus, points 2 and three lines in the same equipotential surface, and hence the potential energy along both paths are the same.

From the law of conservation of energy, the total initial energy of the system is equal to the system's total final energy. Thus, both the protons have the same kinetic energy and hence the same velocity.