An electron (charge -e ) is projected horizontally into the...

An electron (charge $-e$ ) is projected horizontally into the space between two oppositely charged parallel plates. The electric field between the plates is $500.0 \mathrm{~N} / \mathrm{C}$ upward. If the vertical deflection of the electron as it leaves the plates has magnitude $3.0 \mathrm{~mm}$, how much has its kinetic energy increased due to the electric field? [Hint: First find the potential difference through which the electron moves.]

Key Concepts

Electric Field

An electric field is a region of space where a charge experiences a force. It is quantified by its magnitude, which tells you how strong the force is per unit charge, and its direction, which indicates the direction of the force on a positive charge. In uniform fields, such as between two parallel plates, the field is constant in magnitude and direction.

Potential Difference

A potential difference is the difference in electric potential (or voltage) between two points in an electric field. In uniform fields, it can be calculated as the product of the electric field strength and the displacement in the direction of the field. This concept is crucial for determining the energy change associated with moving a charge between those points.

Work Done on a Charged Particle

When a charged particle moves through an electric field, the electric force does work on the particle. This work is given by the product of the charge and the potential difference through which it moves. The work done results in a change in the particle’s kinetic energy, according to the work-energy principle.

Energy Conversion

The increase in the kinetic energy of a charged particle in an electric field comes from the conversion of electric potential energy into kinetic energy. As the particle moves from one potential to another, the energy provided by the work done by the field is manifested as an increase in its motion energy.

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