Question

two large parallel conducting plates are separated by a distance d, placed in a vacuum, and connected to a source of potential difference v. an oxygen ion, with charge -2e, starts from rest on the surface of one plate and accelerates to the other. if e denotes the magnitude of the electron charge, the final kinetic energy of this ion is
○ ev/2
○ 2ev
○ vd/e
○ ev/d
○ evd
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Explanation:

Step1: Recall work - energy theorem

The work - energy theorem states that the work done on a charged particle is equal to its change in kinetic energy. The work done on a charged particle $q$ moving through a potential difference $V$ is given by $W = qV$.

Step2: Identify the charge of the oxygen ion

The charge of the oxygen ion is $q=- 2e$.

Step3: Calculate the work done on the ion

The work done on the oxygen ion as it moves through the potential difference $V$ is $W=qV=(-2e)\times(-V) = 2eV$ (the negative sign of the charge and the potential difference cancel out in terms of the magnitude of the work done). Since the ion starts from rest, its initial kinetic energy $K_{i}=0$. According to the work - energy theorem $W=\Delta K=K_{f}-K_{i}$. Since $K_{i} = 0$, the final kinetic energy $K_{f}=W$.

Answer:

$2eV$