Question

A long hollow conductor has a rectangular cross section, with sides \(a\) and \(2 a\). One side of length \(2 a\) is charged to a potential \(V_{0}\), and the other three sides are grounded \((V=0)\). (a) By using a conformal transformation, find \(\sigma\), the charge density at the point \(P\), the midpoint of its side, and also find the total charge per unit length on that side. (b) Use separation of variable techniques on Laplace's equation to solve the same problem.

Short answer

Use conformal transformation and Laplace's equation to find potentials. Calculate surface charge density at midpoint and total charge per unit length.

Step-by-step solution

Set up the Problem

Define the parameters: the rectangular cross section has sides with lengths \(a\) and \(2a\). One side of length \(2a\) is charged to a potential \(V_{0}\), and the other three sides are grounded (\(V=0\)). The potential inside the conductor follows Laplace's equation.

Use a Conformal Transformation

Apply a conformal transformation to map the rectangle to a different geometry that is easier to solve. Select a transformation that simplifies the boundary conditions, such as mapping the rectangle to the upper half-plane.

Find the Potential

Solve for the potential \(V\) in the transformed plane using the known boundary conditions. Use the known conformity relations and inverse transformations to get back to the potential in the original configuration.

Calculate Surface Charge Density

From the potential, calculate the electric field \(E\) near the surface using the relation \(E = -abla V\). The surface charge density \(\sigma\) at point \(P\) (the midpoint of the charged side) can then be found using \(\sigma = \epsilon_0 E\). The point \(P\) corresponds to \((a,0)\) in the original configuration.

Determine Total Charge per Unit Length

Integrate the surface charge density \(\sigma\) over the entire charged side to find the total charge per unit length \(Q_{L}\). This involves integrating from \(0\) to \(2a\).

Verify Solution Using Separation of Variables

Set up Laplace's equation, \(abla^2 V = 0\), with the given boundary conditions \(V(0,y)=0\), \(V(a,y)=0\), \(V(x,0)=0\), and \(V(x,2a)=V_0\). Use separation of variables \(V(x,y)=X(x)Y(y)\). Solve the resulting ordinary differential equations, and apply boundary conditions to find the potential function \(V(x,y)\).

Complete the Solution

From the calculated potential \(V(x,y)\), find the electric field and surface charge density as described in Step 4. Compare and match results to ensure consistency between both methods.

Key concepts

Laplace's Equation

Laplace's equation is a second-order partial differential equation commonly used in physics, especially in electromagnetism. It is written as \(abla^2 V = 0\). This equation describes the behavior of scalar potential fields in a charge-free space.
Typically, Laplace's equation comes into play when we're looking at the electric potential in regions where there are no charges. Since the divergence of the electric field \( abla \cdot E = 0 \) in such regions and \( E = - abla V \), applying the divergence to both sides leads us directly to Laplace’s equation.

Separation of Variables

Separation of variables is a powerful technique used to solve partial differential equations like Laplace's equation. It involves assuming that the solution can be written as the product of functions, each depending only on one of the variables. For example, in our problem, we assume \( V(x,y) = X(x)Y(y) \).
This splits the original PDE into two ODEs, which can be solved individually. Once you solve these, you combine the results to find the general solution for the potential function. This method works well with boundary conditions, providing a clear and systematic way to find solutions.

Electric Potential

Electric potential \( V \) is a scalar quantity representing the potential energy per unit charge at a point in an electric field. It's a vital concept in electromagnetism and plays a crucial role in solving problems involving electric fields and potentials.
In our exercise, one side of the rectangle is charged to a potential \( V_0 \), while the other sides are grounded \( (V=0)\). Calculating the potential at every point within the rectangle, using methods like conformal transformation or separation of variables, is key to finding other quantities like electric field near the surface and surface charge density.

Surface Charge Density

Surface charge density \( \sigma \) denotes the amount of charge per unit area on a conductive surface. It is connected to the electric field just outside the surface by Gauss' law. Specifically, for a conductor, \( \sigma = \epsilon_0 E \) where \(E \) is the electric field perpendicular to the surface.
In the context of our problem, once the potential distribution within the rectangle is known, we can find the electric field by taking the gradient \( E = - abla V \). The surface charge density at any point is then calculated directly from the electric field value at that point.

Boundary Conditions in Electromagnetism

Boundary conditions are crucial for solving differential equations like Laplace’s equation because they define the behavior of solutions at the edges of the domain. In electromagnetism, these conditions can include fixed potentials, known charges, or specific field values at boundaries.
In our problem, the boundary conditions are set by one side of the rectangle being at potential \( V_0 \), and the other three sides being grounded (\( V=0 \)). These conditions must be applied to the general solution obtained from separation of variables or conformal transformation to find the specific potential function for this setup.