Question

An electron is projected horizontally at a speed of \(1.2 \times\) \(10^{7} \mathrm{~m} / \mathrm{s}\) into an electric field that exerts a constant vertical force of \(4.5 \times 10^{-16} \mathrm{~N}\) on it. The mass of the electron is \(9.11 \times 10^{-31} \mathrm{~kg} .\) Determine the vertical distance the electron is deflected during the time it has moved forward \(33 \mathrm{~mm}\) horizontally.

Short answer

The vertical distance the electron is deflected is approximately 0.935 mm.

Step-by-step solution

Calculate the time of flight

First, calculate the time it takes for the electron to travel 33 mm. Since the horizontal motion is constant and there are no horizontal forces acting on the electron, use the formula \(time = \frac{distance}{speed}\) . So the time \(t\) is given by: \( t = \frac{33 mm}{1.2 \times 10^{7} m/s} = 2.75 \times 10^{-9} s \)

Compute the acceleration

Next, compute the acceleration of the electron in the vertical direction due to the electric field. The acceleration \(a\) due to a force \(F\) is given by Newton's second law, \( F = m \cdot a \), where \( m \) is the mass of the electron. So we can say that \( a = \frac{F}{m} = \frac{4.5 \times 10^{-16} N}{9.11 \times 10^{-31} kg} = 4.94 \times 10^{14} m/s^2 \)

Calculate vertical displacement

Finally, calculate the vertical distance the electron is deflected. Using the equation for the displacement under constant acceleration, \(y = \frac{1}{2} a t^2\), where \( y \) is the vertical displacement, \( a \) is the acceleration, and \( t \) is the time. Plugging the earlier calculated values into this equation, we get: \(y = \frac{1}{2} \times 4.94 \times 10^{14} m/s^2 \times (2.75 \times 10^{-9} s)^2 = 0.935 mm \).

Key concepts

Electric Force

Electric force plays a crucial role in explaining how particles like electrons behave in electric fields. It is a fundamental interaction between charged objects and is described by Coulomb’s law. In the scenario where an electron is projected into an electric field, the electric force is the influence that causes it to accelerate.

To be specific, the force experienced by a charge in an electric field (in this case, an electron), can be calculated using the equation: \( F = qE \), where \( q \) is the charge of the electron and \( E \) is the electric field strength. However, in our exercise, the electric field exerts a constant vertical force given directly as \( 4.5 \times 10^{-16} \mathrm{~N} \), therefore we don't need to calculate it from the charge and field strength.

Understanding the magnitude and direction of this force is essential as it determines the electron's subsequent motion. Since force is a vector quantity, it has both magnitude and direction, and the direction of the force is what dictates the vertical displacement of the electron.

Electron Acceleration

When an electron is subjected to an electric force, it experiences acceleration. According to Newton's second law, the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass (\( F = m \cdot a \) ). This relationship is vital to understanding how we calculate the electron's acceleration in our problem.

In the electron's case, its mass is \( 9.11 \times 10^{-31} \mathrm{~kg} \) and the electric force applied is \( 4.5 \times 10^{-16} \mathrm{~N} \). Using Newton's second law, we find acceleration by rearranging the formula to \( a = \frac{F}{m} \) and substituting in the given values. This gives us \( a = \frac{4.5 \times 10^{-16} N}{9.11 \times 10^{-31} kg} = 4.94 \times 10^{14} m/s^2 \).

This tells us how quickly the velocity of the electron changes as it moves through the electric field. A greater force or a smaller mass would result in a greater acceleration. This acceleration is constant, an important feature when it comes to using kinematic equations to describe the electron's motion.

Kinematic Equations

Kinematic equations enable us to link the motion of an object with its initial velocity, constant acceleration, and time to calculate various motion parameters such as displacement. These formulas are a cornerstone of classical mechanics, particularly when analyzing objects moving under constant acceleration.

In this exercise, we used the kinematic equation for vertical displacement under constant acceleration, given by \( y = \frac{1}{2} a t^2 \) where \( y \) is the vertical displacement, \( a \) is the acceleration, and \( t \) is the time. The specific equation is utilized because, in this scenario, the initial vertical velocity of the electron is zero; hence, the term involving initial velocity disappears, simplifying our calculations.

By plugging in the acceleration (\( 4.94 \times 10^{14} m/s^2 \) ) and the time of flight (\( 2.75 \times 10^{-9} s \) ) into the equation, we could determine the vertical displacement to be \( 0.935 mm \). It’s important to note that if the electron had an initial vertical velocity or if there were additional forces acting on it (like gravity), we would need to use more comprehensive kinematic equations to solve for its motion.