Problem 34 - College Physics

Question

A particle with a mass of \(2.0 \times 10^{-5} \mathrm{~kg}\) and a charge of \(+2.0 \mu \mathrm{C}\) is released in a (parallel plate) uniform horizontal electric field of \(12 \mathrm{~N} / \mathrm{C}\). (a) How far horizontally does the particle travel in \(0.50 \mathrm{~s} ?\) (b) What is the horizontal component of its velocity at that point? (c) If the plates are \(5.0 \mathrm{~cm}\) on each side, how much charge is on each?

Step-by-Step Solution

Answer

(a) 0.15 m, (b) 0.60 m/s, (c) 2.66 x 10^-13 C on each plate.

Step 1: Calculate the Force on the Particle

The electric force on the particle can be calculated using the formula \( F = qE \), where \( q \) is the charge and \( E \) is the electric field. Substitute the given values to find the force:\[F = (2.0 \times 10^{-6} \, \mathrm{C})(12 \, \mathrm{N/C}) = 2.4 \times 10^{-5} \, \mathrm{N}\]

Step 2: Calculate the Acceleration of the Particle

Use Newton's second law, \( F = ma \), to find the acceleration \( a \) of the particle, where \( m \) is the mass. Rearrange the formula to solve for \( a \):\[a = \frac{F}{m} = \frac{2.4 \times 10^{-5} \, \mathrm{N}}{2.0 \times 10^{-5} \, \mathrm{kg}} = 1.2 \, \mathrm{m/s^2}\]

Step 3: Determine the Horizontal Distance Traveled

The particle starts from rest, so use the equation of motion \( s = ut + \frac{1}{2}at^2 \) to find the distance \( s \), where \( u = 0 \) is the initial velocity, \( a \) is the acceleration, and \( t = 0.50 \, \mathrm{s} \) is the time.\[s = 0 + \frac{1}{2}(1.2 \, \mathrm{m/s^2})(0.50 \, \mathrm{s})^2 = 0.15 \, \mathrm{m}\]

Step 4: Calculate the Horizontal Velocity at 0.50 s

Use the basic kinematic equation \( v = u + at \) to calculate the velocity \( v \) after \( 0.50 \, \mathrm{s} \). Since it starts from rest, \( u = 0 \).\[v = 0 + (1.2 \, \mathrm{m/s^2})(0.50 \, \mathrm{s}) = 0.60 \, \mathrm{m/s}\]

Step 5: Calculate Surface Charge Density on Plates

If the plates are \(5.0 \, \mathrm{cm} = 0.05 \, \mathrm{m} \) on each side, calculate the area \( A \) of one plate: \( A = (0.05 \, \mathrm{m})^2 = 2.5 \times 10^{-3} \, \mathrm{m^2} \). The charge on each plate can be found using the relation \( E = \frac{\sigma}{\varepsilon_0} \), where \( \sigma \) is the surface charge density, and \( \varepsilon_0 \approx 8.85 \times 10^{-12} \, \mathrm{C^2/(N \, m^2)} \) is the permittivity of free space.\[\sigma = E \varepsilon_0 = (12 \, \mathrm{N/C})(8.85 \times 10^{-12} \, \mathrm{C^2/(N \, m^2)}) = 1.062 \times 10^{-10} \, \mathrm{C/m^2}\]

Step 6: Calculate the Charge on Each Plate

Multiply the surface charge density \( \sigma \) by the area \( A \) to find the charge on each plate:\[Q = \sigma A = (1.062 \times 10^{-10} \, \mathrm{C/m^2})(2.5 \times 10^{-3} \, \mathrm{m^2}) = 2.655 \times 10^{-13} \, \mathrm{C}\]

Key Concepts

Particle Motion

Understanding the motion of a particle in an electric field helps describe its path and velocity. In this context, a particle with a specific mass and charge is released in a uniform electric field. The electric field influences the particle's motion directly, causing it to accelerate uniformly. When analyzing the motion, it's essential to know that it starts from rest. This means it initially has zero velocity. As time progresses, thanks to the constant electric force acting on it, the particle speeds up and covers more distance. This change in motion can be predicted and calculated using the principles of physics, specifically under the umbrella of kinematics and dynamics.

When we discuss particles in electric fields, assuming no other forces act on the particle (like gravity or air resistance), its horizontal path and acceleration can be neatly predicted. The concept of particle motion forms a foundation for understanding more complex systems where multiple forces might act simultaneously.

Electric Force

Electric force is a fundamental concept when studying particles in electric fields. The force exerted on a charged particle in an electric field is calculated using the equation: \[ F = qE \] where \( F \) is the force, \( q \) is the charge, and \( E \) is the electric field strength. This force is vectorial, meaning it has both magnitude and direction.

In the given exercise, a particle with a charge of \(+2.0 \, \mu \mathrm{C}\) experiences a force due to the uniform electric field of \(12 \, \mathrm{N/C}\). The resulting force points in the direction of the field, accelerating the particle. It's important to note that this force results in constant acceleration as long as the field remains uniform.

• This approach demonstrates how charges interact with electric fields naturally by exerting forces.
• These forces are responsible for driving the motion studied in electromagnetic field theory.

Understanding electric forces is crucial for exploring how particles move within electric fields.

Newton's Second Law

Newton's second law of motion is a foundational principle used to relate the motion of a particle with the forces acting on it. It is represented by the equation: \[ F = ma \] where \( m \) is the mass of the object, and \( a \) is its acceleration. Through this law, we learn that the force acting on an object is equal to the mass of that object multiplied by its acceleration. This concept helps decipher the dynamics of how and why a particle accelerates in response to forces, such as those from an electric field.

In our example, knowing the mass of the particle (\(2.0 \times 10^{-5} \, \mathrm{kg}\)) and the force it experiences (which we previously calculated), we can determine its acceleration using \[ a = \frac{F}{m} \] This step bridges the gap between understanding the electric force and predicting the resultant motion through calculated acceleration. Furthermore, with constant force and unchanging mass, the acceleration remains constant, simplifying analysis and predictions.

Kinematic Equations

Kinematic equations provide a framework for understanding the motion of objects, specifically their displacement, velocity, time, and acceleration relationships. These equations assume constant acceleration, which matches our conditions since the force derived from an electric field results in uniform acceleration.

One key equation is \[ s = ut + \frac{1}{2}at^2 \] where \( s \) is displacement, \( u \) is initial velocity, \( a \) is acceleration, and \( t \) is time. In the exercise, since the particle begins at rest, \( u = 0 \), simplifying calculations. This equation allows us to find how far the particle travels in the given time.

• The horizontal distance the particle covers in a specific timeframe is determined using this formula.
• Such equations reveal insights into the particle's trajectory.

Another notable kinematic equation is \[ v = u + at \] which finds the particle's velocity after a certain time, considering initial velocity and acceleration due to the electric field. These tools are indispensable for students in predicting and analyzing particle motion comprehensively.