Problem 80 - University Physics with Modern Physics

Question

In a region where there is a uniform electric field that is upward and has magnitude 3.60 \(\times 10^4 \space N/\)C, a small object is projected upward with an initial speed of 1.92 m\(/\)s. The object travels upward a distance of 6.98 cm in 0.200 s. What is the object's charge-to-mass ratio \(q/m\)? Assume \(g = 9.80 \space m/s^2\), and ignore air resistance.

Step-by-Step Solution

Answer

The object's charge-to-mass ratio is approximately \(-1.67 \times 10^{-4} \, \text{C/kg}\).

Step 1: Compute Forces on the Object

The object experiences two forces: the gravitational force \( F_{gravity} = mg \) and the electric force \( F_{electric} = qE \). As the object is moving upwards, it faces a net force of \( F_{net} = qE - mg \) in the upward direction.

Step 2: Apply Newton's Second Law

According to Newton's Second Law, \( F_{net} = ma \), where \(a\) is the acceleration. Substituting the net force gives us: \[ma = qE - mg \] Rearranging terms yields: \[a = \frac{q}{m}E - g \] Thus, the acceleration \(a\) is expressed in terms of \(\frac{q}{m}\).

Step 3: Calculate Acceleration Using Kinematics

Using the kinematic equation, \(s = ut + \frac{1}{2}at^2\), where \(s\) is the displacement, \(u\) is the initial velocity, and \(t\) is the time. Plugging in the values: \[0.0698 \, m = 1.92 \, \text{m/s} \times 0.200 \, s + \frac{1}{2}a(0.200 \, s)^2 \] This simplifies to: \[a \times 0.0200 \, s^2 = 0.0698 \, m - 0.384 \, m \] \[a = \frac{0.0698 \, m - 0.384 \, m}{0.0200 \, s^2} \] Solving gives \(a = -15.8 \, \text{m/s}^2\).

Step 4: Solve for Charge-to-Mass Ratio

We have the expression for acceleration in Step 2: \[a = \frac{q}{m}E - g\]. Substitute \(a\) and solve for \(\frac{q}{m}\): \[-15.8 = \frac{q}{m} \times 3.60 \times 10^4 - 9.80 \] \[\frac{q}{m} = \frac{-15.8 + 9.80}{3.60 \times 10^4}\]Solving the equation gives \( \frac{q}{m} \approx -1.67 \times 10^{-4} \, \text{C/kg}\).

Key Concepts

Charge-to-Mass Ratio

The charge-to-mass ratio, often denoted as \( \frac{q}{m} \), is a crucial parameter in physics when studying particles in electric fields. It essentially measures how much charge is present per unit mass of a particle. This ratio is significant because it helps us understand how a particle will move under the influence of an electric field.

• A higher charge-to-mass ratio means that, for a given electric field, the particle experiences more force per unit of mass. This can lead to larger accelerations.
• Conversely, a lower charge-to-mass ratio means the particle experiences less force, resulting in smaller accelerations under the same field conditions.

In the given problem, we're tasked to find the charge-to-mass ratio of a small object in an electric field. The force due to the electric field is \( F_{electric} = qE \) and the gravitational force is \( F_{gravity} = mg \). Seeing both act on the particle and knowing the net force direction helps us set up the equations to find \( \frac{q}{m} \). Since the problem provides specifics about the object's movement and the forces acting upon it, this allows a detailed calculation using these two forces.

Kinematics

Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. It helps us describe how an object moves, and in this problem, it is used to calculate the acceleration of the small object.

• We applied the equation for uniformly accelerated motion, \( s = ut + \frac{1}{2}at^2 \), which links displacement \(s\), initial speed \(u\), acceleration \(a\), and time \(t\).
• Given: the object moves a displacement of 6.98 cm (converted to meters as 0.0698 m) with an initial speed of 1.92 m/s over a time period of 0.200 s.

By substituting all these values into the kinematic equation, we can solve for the object's acceleration \(a\). This calculated acceleration helps us further solve for other unknowns, such as the charge-to-mass ratio in the provided equations. Kinematics gives us a backdrop to quantify motion in terms of measurable factors like displacement and time, leading us to derive useful parameters like acceleration.

Newton's Second Law

Newton's Second Law is a cornerstone of classical mechanics and states that the force acting on an object is equal to the mass of that object times its acceleration, expressed as \( F = ma \). This law provides a direct relationship between the force applied to an object, its mass, and its resultant acceleration.

• In this exercise, Newton's Second Law is instrumental in understanding the object's response to combined forces due to gravity and an electric field.
• The net force \( F_{net} \) is derived from both the electric force \( qE \) and the gravitational force \( mg \), thus \( F_{net} = qE - mg \). This reflects the object's upward movement against gravity.

By linking \( F_{net} \) to \( ma \), we rearrange to find the acceleration \( a \) as \( a = \frac{q}{m}E - g \). This way, after calculating \( a \) using kinematics, we can isolate \( \frac{q}{m} \) and solve for it. Newton's Second Law teaches us how forces and motion are inter-related and lays the groundwork for calculating different physical properties such as the charge-to-mass ratio in this scenario.