Question

Let us assume that we have a very thin, square, imperfectly conducting plate \(2 \mathrm{~m}\) on a side, located in the plane \(z=0\) with one corner at the origin such that it lies entirely within the first quadrant. The potential at any point in the plate is given as \(V=-e^{-x} \sin y .(a)\) An electron enters the plate at \(x=0, y=\pi / 3\) with zero initial velocity; in what direction is its initial movement? ( \(b\) ) Because of collisions with the particles in the plate, the electron achieves a relatively low velocity and little acceleration (the work that the field does on it is converted largely into heat). The electron therefore moves approximately along a streamline. Where does it leave the plate and in what direction is it moving at the time?

Short answer

#Question# Determine the initial direction of the electron's movement when it enters an imperfectly conducting plate at x=0 and y=π/3 if the potential function within the plate is given by V = -e^(-x) sin(y). #Answer# The initial direction of the electron's movement is given by the angle θ = arctan((-q * cos(π/3)) / (q * sin(π/3))).

Step-by-step solution

Find the electric field using the potential function

The potential function V is given by: V = -e^(-x) sin(y). To find the electric field E within the plate, we can use the gradient of V: E = -∇V. Therefore, we need to take the partial derivatives of V with respect to x and y. Ex = -(dV/dx) = -(-e^(-x) sin(y)) = e^(-x) sin(y) Ey = -(dV/dy) = -(e^(-x) cos(y))

Calculate the force on the electron when it enters the plate

The force, F, on the electron is given by the charge of the electron (q) multiplied by the electric field at the point where it enters the plate (x=0, y=π/3). So, we can calculate Fx and Fy components as: Fx = q * Ex = q * e^(0) * sin(π/3) = q * sin(π/3) Fy = q * Ey = q * (-e^(0) * cos(π/3)) = -q * cos(π/3)

Determine the initial direction of the electron's movement

The initial direction of the electron's movement is given by the arctangent of the ratio of the force components: θ = arctan(Fy / Fx) = arctan((-q * cos(π/3)) / (q * sin(π/3))) The direction of movement is given by the angle θ.

Approximate the path of the electron by following the path of a streamline

Since the electron moves approximately along a streamline, we can compute the streamline function from the electric field components: Streamline function: ψ(x, y) = ∫(Ey)dx - ∫(Ex)dy ψ(x, y) = ∫(-e^(-x) cos(y)dx) - ∫(e^(-x) sin(y)dy) ψ(x, y) = e^(-x) cos(y) + e^(-x) cos(y) + C To find the constant C, we can use the initial position of the electron, x=0 and y=π/3: ψ(0, π/3) = 1/2 + 1/2 + C = 0 C = -1 So the streamline function is: ψ(x, y) = e^(-x) cos(y) + e^(-x) cos(y) - 1

Determine where the electron leaves the plate and its direction

The electron leaves the plate when one of the coordinates x or y is equal to 2, as the plate has a side length of 2 m. In order to find the direction of the electron's movement at that time, we can calculate the tangent of the angle θ at the point where the electron leaves the plate. To identify the point, we must find the values of x and y that satisfy the equation ψ(x, y) for x=2 or y=2. From this relation and the angle θ calculated in Step 3, we can apprehend the final direction of the movement of the electron within the plate.

Key concepts

Electric Field

The electric field is a fundamental concept in electromagnetism, representing the force felt by a charged particle in the presence of another charge or a potential. In this case, the electric field is derived from a potential function, which is specified as \( V = -e^{-x} \sin(y) \), across a two-dimensional plane. To find the electric field \( E \), we use the gradient of the potential function: \( E = -abla V \). This involves calculating the partial derivatives of \( V \) concerning \( x \) and \( y \), which respectively yield \( E_x = e^{-x} \sin(y) \) and \( E_y = -e^{-x} \cos(y) \). Each component describes how the electric field affects electron movement in the respective direction. Understanding these components is key for predicting how the electron will initially accelerate as it enters the electric field.

Potential Function

The potential function determines how the electric field behaves within the plate. In essence, the function \( V = -e^{-x} \sin(y) \) describes the energy landscape of the field, which is crucial for understanding how an electron will move. This potential is particularly interesting because it combines exponential decay \(-e^{-x}\) and sinusoidal behavior \(\sin(y)\), suggesting variation in both directions across the plane.

• Exponential component: Indicates the potential's rapid decrease in the x-direction.
• Sinusoidal component: Reflects oscillating potential changes along the y-direction.

The potential function's complexity reflects that the electron will experience combined influences in x and y directions, leading to specific trajectories that change as it traverses the plate.

Electron Movement

When the electron first enters the thin conductive plate with coordinates \( x = 0, y = \pi/3 \), its movement is governed by the electrical forces acting on it. These forces are derived from the earlier calculated electric field components: \( F_x = q \cdot \sin(\pi/3) \) and \( F_y = -q \cdot \cos(\pi/3) \), where \( q \) represents the electron's charge. These forces dictate the electron's initial direction.
The angle of this movement, \( \theta \), is found using the arctangent of the force ratio:
\( \theta = \arctan \left( \frac{-q \cdot \cos(\pi/3)}{q \cdot \sin(\pi/3)} \right) \).
The simplification of the electron's charge allows us to find \( \theta \) with ease, indicating that the direction continues to lean towards the plate's streamline pathway due to the low velocity and acceleration from collisions.

Streamline

In the context of this problem, a streamline describes a path that is consistent with the direction of the electric field, and thus the electron's movement. Because the electron's velocity is low, its path can be approximated by a streamline. This path can be calculated from the electric field using the expression:

\( \psi(x, y) = \int (E_y) \, dx - \int (E_x) \, dy \)

For this exercise, it is

\( \psi(x, y) = e^{-x} \cos(y) + e^{-x} \cos(y) - 1 \).

• Streamlines indicate the path that would show the trajectory of the field along which the electron moves.
• For this plate, the streamlines lie in a way that ensures the electron smoothly transitions out of the plate at its boundary.

Understanding streamlines helps predict where the electron will exit without needing to observe complex dynamics directly.

Gradient

The gradient of a potential function is used to determine the electric field, influencing how charged particles like electrons move. Specifically, the gradient \( abla V \) for our potential function \( V = -e^{-x} \sin(y) \) is:

• Partial derivative with respect to x: \( \frac{\partial V}{\partial x} = e^{-x} \sin(y) \)
• Partial derivative with respect to y: \( \frac{\partial V}{\partial y} = -e^{-x} \cos(y) \)

These derivatives define the direction and magnitude of the electric field changes. The negative sign in \( E = -abla V \) indicates the direction opposite to increasing field strength, guiding the initial movement of the electron within the plane. The gradient shows us not just the most direct path of force application but also an area of declining potential energy for the moving electron.