Problem 74 - University Physics with Modern Physics

Question

CP At \(t=0\) a very small object with mass 0.400 \(\mathrm{mg}\) and charge \(+9.00 \mu \mathrm{C}\) is traveling at 125 \(\mathrm{m} / \mathrm{s}\) in the \(-x\) -direction. The charge is moving in a uniform electric field that is in the +y-direction and that has magnitude \(E=895 \mathrm{N} / \mathrm{C}\) . The gravitational force on the particle can be neglected. How far is the particle from the origin at \(t=7.00 \mathrm{ms} ?\)

Step-by-Step Solution

Answer

The particle is approximately 1.005 m from the origin at 7.00 ms.

Step 1: Understand the Problem

We have a charged particle moving initially in the \(-x\) direction with a uniform electric field acting in the \(+y\) direction. We need to find the distance of the particle from its starting point (the origin) at \(t = 7.00 \,\mathrm{ms}\). We neglect gravitational forces.

Step 2: Analyze Initial Conditions

The initial velocity \(v_{0x} = -125 \, \mathrm{m/s}\) in the \(-x\) direction. The object has a mass of \(0.400 \,\mathrm{mg} = 0.400 \times 10^{-6} \, \mathrm{kg}\) and a charge of \(+9.00 \, \mu\mathrm{C} = 9.00 \times 10^{-6} \, \mathrm{C}\). The electric field \(E = 895 \, \mathrm{N/C}\) is in the \(+y\) direction.

Step 3: Determine Electric Force

The electric force \(F\) acting on the particle in the \(+y\) direction is given by \(F = qE\), where \(q = 9.00 \times 10^{-6} \, \mathrm{C}\) and \(E = 895 \, \mathrm{N/C}\). Substituting the values, \(F = 9.00 \times 10^{-6} \, \mathrm{C}\times 895 \, \mathrm{N/C} = 8.055 \times 10^{-3} \, \mathrm{N}\).

Step 4: Determine Acceleration in y-direction

Use Newton's second law \(F = ma\) to find the acceleration \(a\) in the y-direction. \(a_y = \frac{F}{m} = \frac{8.055 \times 10^{-3} \, \mathrm{N}}{0.400 \times 10^{-6} \, \mathrm{kg}}\).This gives \(a_y = 20137.5 \, \mathrm{m/s^2}\).

Step 5: Find Displacement in x-direction

The displacement in the x-direction after time \(t = 7.00 \, \mathrm{ms} = 7.00 \times 10^{-3} \, \mathrm{s}\) can be calculated using \(x = v_{0x}t\).Substitute \(v_{0x} = -125 \, \mathrm{m/s}\) and \(t = 7.00 \times 10^{-3} \, \mathrm{s}\), we get \(x = -125 \, \mathrm{m/s} \times 7.00 \times 10^{-3} \, \mathrm{s} = -0.875 \, \mathrm{m}\).

Step 6: Find Displacement in y-direction

The displacement in the y-direction is given by \(y = \frac{1}{2} a_{y} t^2\).Substitute \(a_y = 20137.5 \, \mathrm{m/s^2}\) and \(t = 7.00 \times 10^{-3} \, \mathrm{s}\)\[ y = \frac{1}{2} \times 20137.5 \, \mathrm{m/s^2} \times (7.00 \times 10^{-3} \, \mathrm{s})^2 \approx 0.4942 \, \mathrm{m} \]

Step 7: Calculate Distance from Origin

The total distance from the origin can be found using the Pythagorean theorem:\[ d = \sqrt{x^2 + y^2} = \sqrt{(-0.875)^2 + (0.4942)^2} \].This results in:\[ d \approx \sqrt{0.765625 + 0.244241} \approx \sqrt{1.009866} \approx 1.005 \mathrm{m} \]

Step 8: Conclusion

Therefore, at \(t = 7.00 \, \mathrm{ms}\), the particle is approximately \(1.005 \, \mathrm{m}\) away from the origin.

Key Concepts

Electric Force

In the world of electric fields, electric force is a critical concept to understand. It is the force exerted by an electric field on a charged object. When a charged particle is placed in an electric field, it experiences a force that can cause it to accelerate. This force is determined by two key factors: the charge of the particle and the strength of the electric field. The formula used to calculate this electric force is given by:

• \( F = qE \)

where \( F \) is the electric force, \( q \) represents the charge of the particle, and \( E \) stands for the magnitude of the electric field.
In our problem, a small object with a charge of

• \( +9.00 \mu \mathrm{C} \)

is placed in an electric field of

• \( 895 \, \mathrm{N/C} \)

This results in an electric force calculated as:

• \( F = 9.00 \times 10^{-6} \, \mathrm{C} \times 895 \, \mathrm{N/C} = 8.055 \times 10^{-3} \, \mathrm{N} \)
.

Charged Particles

Charged particles are objects that have an excess or deficit of electrons, giving them a net electric charge. These charges can either be positive or negative. Positive charges have more protons than electrons, while negative charges have more electrons than protons.
When charged particles move through electric fields, they experience forces that depend on their charge and the field's strength. In physics, analyzing the movement and interaction of such particles is crucial because it helps us understand and predict their behaviors in different scenarios.
In this scenario, our particle is positively charged with

• \( +9.00 \mu \mathrm{C} \)

which interacts with the electric field, moving initially in the

• ext{-x} direction at a speed of
  • \(125 \, \mathrm{m/s} \)
  .

Kinematics

Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. When we talk about motion, we often refer to quantities like displacement, velocity, and acceleration.
In our problem, kinematics helps us determine how far the particle travels both in the x-direction and y-direction within a given time. We use the initial velocity and the time elapsed for motion along the x-axis, calculated as:

• \( x = v_{0x}t = -125 \, \mathrm{m/s} \times 7.00 \times 10^{-3} \, \mathrm{s} = -0.875 \, \mathrm{m} \)

For the y-direction, the kinematic equation includes acceleration derived from electric force, resulting in the formula:

• \( y = \frac{1}{2} a_{y} t^2 \)

Substituting the values gives us the y-direction displacement:

• \( y \approx 0.4942 \, \mathrm{m} \).

Newton's Second Law

Newton's second law of motion is a fundamental principle that connects the concepts of force, mass, and acceleration. This law states that the force acting on an object is equal to the mass of that object multiplied by its acceleration:

• \( F = ma \)

In this problem, we use Newton's second law to find the acceleration of the particle in the y-direction. The electric force exerted on the charged particle causes it to accelerate, and this acceleration can be found by rearranging the formula to:

• \( a_y = \frac{F}{m} \)

Using our electric force \( F = 8.055 \times 10^{-3} \, \mathrm{N} \) and the mass \( m = 0.400 \times 10^{-6} \, \mathrm{kg} \), we calculate:

• \( a_y = \frac{8.055 \times 10^{-3} \, \mathrm{N}}{0.400 \times 10^{-6} \, \mathrm{kg}} = 20137.5 \, \mathrm{m/s^2} \)

This acceleration allows us to determine the y-direction motion of the particle.