91Ó°ÊÓ

Electrons are subatomic particles with a negative charge. They are very light and often participate in electrical phenomena. When discussing electron motion, especially within an electric field, it's important to understand that their motion is significantly affected by external forces. Here, the electron initially moves at a speed of \(7.5 \times 10^5 \text{ m/s}\). Due to its charge, an electric field can exert a force on the electron, changing its velocity and eventually bringing it to rest if the field is in the right direction.
The electron motion analysis importantly involves both the initial and final states of speed and position, setting up the stage for deeper investigations using other principles like kinematics and dynamics.

Kinematic equations are used to describe the motion of bodies under constant acceleration. In this problem, one useful kinematic equation is \(v_f^2 = v_i^2 + 2a \times d\). This equation relates the final velocity \(v_f\), initial velocity \(v_i\), acceleration \(a\), and distance \(d\) over which the acceleration occurs.
Specific to our exercise, we set \(v_f = 0\) since the electron must come to a stop, and \(v_i = 7.5 \times 10^5 \text{ m/s}\). The stopping distance \(d\) is \(0.04 \text{ m}\). By rearranging this equation, we can solve for the electron's acceleration \(a\) needed to stop it over the given distance.

Newton’s Second Law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration: \(F = ma\). This principle helps us understand how forces result in changes in motion or acceleration.
In the context of our exercise, after calculating the electron's acceleration, we use its mass \(m = 9.11 \times 10^{-31} \text{ kg}\) with the acceleration to find the force \(F\) acting on the electron. Understanding this force is critical since it directly connects to the electric force experienced by the electron due to the electric field.

Electric force is the force exerted by an electric field on a charged particle. It is given by the equation \(F = eE\), where \(F\) is the force, \(e = 1.6 \times 10^{-19} \text{ C}\) is the charge of the electron, and \(E\) is the electric field strength.
In our scenario, the electric force must match the opposing force calculated using Newton's second law for the electron to stop as desired. By equating \(ma\) to \(eE\), we derive the expression for the electric field strength \(E = \frac{ma}{e}\). Substituting in the known values of mass, acceleration, and electron charge allows us to solve for the needed electric field strength.