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?

Step by step solution

01

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))

02

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)

03

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.

04

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

05

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.

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