91Ó°ÊÓ

Kinematic equations form the backbone of motion analysis without considering forces directly. They allow us to connect various aspects of an object's movement—such as displacement, initial velocity, acceleration, and time. In this scenario, the equation of interest is:\[ s = ut + \frac{1}{2} a t^2 \]where:

• \(s\) is the displacement (4.50 m in this case),
• \(u\) is the initial velocity (0 m/s, as the electron starts from rest),
• \(a\) is the acceleration,
• \(t\) is the time duration (3.00 μs or \(3.00 \times 10^{-6}\) s).

These equations help deduce the acceleration, which is fundamental in understanding the forces acting on the electron as it moves through the electric field.

Understanding acceleration is key in evaluating how an object’s velocity changes over time under the influence of different forces. In this problem, acceleration is particularly significant because it tells us how rapidly the electron’s velocity changes due to the electric field.The kinematic equation for displacement already sets the stage for calculating acceleration. By substituting the known values into:\[ s = ut + \frac{1}{2} a t^2 \]we isolate and solve for \(a\), resulting in:\[ a = \frac{2s}{t^2} \]Substitute: \(s = 4.50 \text{ m}\), \(t = 3.00 \times 10^{-6} \text{ s}\) to find the acceleration:\[ a = \frac{2 \times 4.50}{(3.00 \times 10^{-6})^2} = 1.00 \times 10^{12} \text{ m/s}^2 \]This calculation shows how quickly the electron responds to the electric field, reinforcing its position as a critical component of this problem.

The electron charge is a fundamental constant and is crucial in calculating the force experienced by the electron in an electric field. The charge of an electron \(e\) is \(1.60 \times 10^{-19} \text{ C}\). This fundamental property plays a direct role in the equation:\[ F = eE \]where \(F\) is the force experienced by the electron and \(E\) is the electric field strength. Using Newton’s second law \(F = ma\), it follows that:\[ eE = ma \]Rearranging gives us the electric field as:\[ E = \frac{ma}{e} \]Using this equation and substituting the known values, the electric field can be calculated. The charge \(e\) of the electron is always negative, affecting how the force is directed relative to the electric field.

The effects of gravity on an electron are often deemed negligible due to its extremely small mass and the dominating influence of the electric field. To quantitatively check this, we calculate the gravitational force \(F_g\) on the electron:\[ F_g = mg \]where \(m = 9.11 \times 10^{-31} \text{ kg}\) (the mass of the electron) and \(g = 9.81 \text{ m/s}^2\) (acceleration due to gravity).This gives a gravitational acceleration equivalent to:\[ a_g = 9.81 \text{ m/s}^2 \]Comparing \(a_g\) to the calculated acceleration from the electric field \(1.00 \times 10^{12} \text{ m/s}^2\), it is evident that gravitational acceleration is many orders of magnitude smaller. Thus, gravity’s effects on the electron’s motion are negligible, ensuring that the primary concern is the dominating electric force driving the electron's acceleration.