Electrostatics
Intermediate
Motion in Electric Field
Source: High school physics (Chinese)
Problem
In a vacuum, a pair of parallel metal plates are separated by 5.6 cm, with a voltage of 50 V applied between them. A doubly-charged oxygen ion is accelerated from rest by the electric field.
What is its kinetic energy when it reaches the other plate?
[Q1] $K_f = 1.6 \times 10^{-17}$ J (or 100 eV)
According to the work-energy theorem, the change in kinetic energy $\Delta K$ is equal to the work $W$ done by the electric field. Since the ion starts from rest, its final kinetic energy $K_f$ is equal to the work done on it.
$$K_f = W = qV$$A doubly-charged ion has a charge of $q = 2e$.
$$K_f = (2e)V$$