91Ó°ÊÓ

The electric field is a fundamental concept in electrostatics. It represents the force exerted on a charge when placed in the vicinity of other electric charges. For a point charge, the electric field is directed away from positive charges and towards negative charges.

In this exercise, we consider an electron in the electric field between two parallel plates. When an electron is placed within this field, it experiences a force due to the electric field. The relation to calculate this force is given by the equation:

• \( F = qE \)

where \( F \) is the force, \( q \) is the charge of the particle (for an electron it is \(-1.6 \times 10^{-19} \mathrm{~C}\)), and \( E \) is the electric field.

• By rearranging we find the field: \( E = \frac{F}{q} \)

In our problem, an electron experiences a force of \(3.9 \times 10^{-15} \mathrm{~N}\), providing us the opportunity to easily calculate the electric field by plugging the values into the formula, resulting in an electric field strength of \( 2.44 \times 10^{4} \mathrm{~N/C} \), only considering its magnitude.

Potential difference, also known as voltage, is the work required to move a charge through an electric field. It is a measure of the potential energy difference between two points in an electric field. This concept is crucial when analyzing systems involving capacitors like in our example.
The potential difference between the two plates can be calculated using the formula:

• \( V = Ed \)

where \( V \) is the voltage, \( E \) is the electric field calculated in the previous step, and \( d \) is the distance separating the plates.

In our exercise, the plates are \(0.12 \mathrm{~m}\) apart. With an electric field of \(2.44 \times 10^{4} \mathrm{~N/C}\), the potential difference across the plates is found by multiplying these two values:

• \( V = 2.44 \times 10^{4} \mathrm{~N/C} \times 0.12 \mathrm{~m} \)
• \( V = 2928 \mathrm{~V} \)

This result tells us how much work is required to move a charge from one plate to the other, highlighting the energy stored in the system.

A parallel plate capacitor is an arrangement of two conducting plates separated by a certain distance. By charging each plate with equal and opposite electric charges, a uniform electric field is created between the plates.

This configuration is significant in many electronic devices for its ability to store and release electrical energy. The uniform electric field between these plates is described by:

• \( E = \frac{V}{d} \)

where \( V \) is the potential difference and \( d \) is the plate separation.
In the exercise given, the plates are 12 cm apart with charges such that an electron placed within experiences a measurable electrostatic force. Capacitors are used to store this electric energy and have many applications, including in computers and telephone systems. By understanding concepts such as electric fields and potential differences in the context of a parallel plate capacitor, students gain insight into how capacitors regulate the flow of electrical currents, which is fundamental to how many electrical devices operate.