An electron is released from rest at the negative plate of a...

An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing). The plates are separated by a distance of 1.2 $\mathrm{cm}$ , and the electric field within the capacitor has a magnitude of 2.1 $\times 10^{6} \mathrm{V} / \mathrm{m}$ . What is the kinetic energy of the electron just as it reaches the positive plate?

An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the
positive plate (see the drawing). The plates are separated by a distance of 1.2 $\mathrm{cm}$ , and the electric field within the capacitor has a magnitude of 2.1 $\times 10^{6} \mathrm{V} / \mathrm{m}$ . What is the kinetic energy of the electron just as it reaches the positive plate?

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Key Concepts

Conversion of Potential Energy to Kinetic Energy

The work-energy principle states that the work done on the electron by the electric field is converted into its kinetic energy. When the electron moves from the negative plate to the positive plate, the change in electric potential energy (given by the product of charge and voltage difference) is equal to the gain in kinetic energy, assuming negligible energy losses.

Voltage Difference

This concept involves the electric potential difference between two points, which in a uniform electric field is given by the product of the electric field strength and the distance over which it acts (V = E × d). The voltage difference across the capacitor plates influences the energy change experienced by the electron.

Electric Field

This concept refers to a region where a charged particle experiences a force. In a capacitor, the electric field is uniform between the plates, and its magnitude determines the force exerted on a charge. The relationship between the field and the force is described by F = qE, where q is the charge of the electron.

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