Question

Find the potential inside a semi-infinite cylindrical conductor, closed at the nearby end, whose cross section is a square with sides of length \(a\). All sides are grounded except the square side, which is held at the constant potential \(V_{0}\).

Short answer

The potential inside the semi-infinite cylindrical conductor is given by \(V(x,y) = \sum_{n=1}^\infty {\frac{2V_{0}}{\sinh(n\pi)} \sin\left(\frac{n\pi y}{a}\right) \sinh\left(\frac{n\pi x}{a}\right)} \)

Step-by-step solution

Formulate the Laplace's equation

Considering a rectangular coordinate system, Laplace's equation in two dimensions (x and y) appears as follows: \(\nabla^{2}V =\frac{\partial^{2}V}{\partial x^{2}} +\frac{\partial^{2}V}{\partial y^{2}} = 0 \). Our task is to solve for the function \(V(x,y)\) which solves the equation.

Apply boundary conditions

We need to apply boundary conditions. Since all sides are grounded, we have \(V|_{y=0} = 0\), \(V|_{x=0} = 0\) and \(V|_{y=a} = 0\), and the ungrounded side is held at the constant potential \(V_{0}\), which gives us \(V_{x=a}=V_{0}\). These boundary conditions will help determine the constants when we solve this equation.

Separate Variables and solve

To solve the Laplace's equation, use the method of separation of variables, stating that the solution can be written as a product of solutions of two ordinary differential equations each in one variable, i.e., \(V(x,y)=X(x)Y(y)\). Substituting this and dividing by \(V(x, y)\) gives \( \frac{X''}{X} + \frac{Y''}{Y} = 0\), leading to two ordinary differential equations \(X''=k^2X\) and \(Y''=-k^2Y\) for some constant \(k^2\), solve with the boundary conditions.

Find the solution

The solutions of the ordinary differential equations are \(X = A\sinh(kx) + B\cosh(kx)\) and \(Y = D\sin(ky)\). Now we apply boundary conditions. On \(x=0\) side, we choose \(B=0\) and from \(x=a\) side we have \(A = V_{0}/\sinh(ka)\). For \(y=0\) side, because sin(0)=0, there is no contributions. For \(y=a\) side we have \(k = n\pi/a \). So combining we have the potential \(V(x,y) = \sum_{n=1}^\infty {\frac{2V_{0}}{\sinh(n\pi)} \sin\left(\frac{n\pi y}{a}\right) \sinh\left(\frac{n\pi x}{a}\right)} \

Key concepts

Boundary Conditions in Electrostatics

Understanding how boundary conditions affect electrostatics is crucial for solving many physics problems, especially those involving electric potential. In the scenario of a semi-infinite cylindrical conductor with a square cross-section, we are given that all sides except one are at zero potential (grounded), and the remaining side is at a constant potential, denoted as \(V_0\). These specifics establish the 'boundary conditions' for the problem.

Mathematically, the boundary conditions lead to restrictions on the potential function \(V(x,y)\) that must satisfy the Laplace equation. For instance, at the grounded boundaries, the potential is zero. This guides in formulating the problem as it gives us concrete values for potential at specific locations. At the square side with potential \(V_0\), we receive another condition that defines the value of the electric potential at that particular boundary. These conditions are not random but stem from physical constraints and are critical for finding a unique solution to the electrostatic problem.

Separation of Variables

The method of 'separation of variables' allows for the simplification of partial differential equations like Laplace's equation in problems involving symmetrical structures such as our cylindrical conductor. By hypothesizing that the two-dimensional potential \(V(x,y)\) can be expressed as the product of two single-variable functions, namely \(X(x)\) and \(Y(y)\), the problem is decomposed into finding solutions for these individual functions.

Through this method, a complex partial differential equation is broken down into simpler ordinary differential equations that can be solved more straightforwardly. The term \(X''/X + Y''/Y = 0\) emerges from the separation, representing two distinct ordinary differential equations with each depending solely on one coordinate. This is beneficial because each equation can now be independently solved using standard methods for ordinary differential equations while respecting the boundary conditions applicable to each coordinate.

Solving Laplace's Equation

Laplace's equation \(abla^{2}V = 0\) is a hallmark of electrostatics and is synonymous with regions of space free of electric charge. When solved, it yields the electric potential \(V\) for a region, given known boundary conditions. In our exercise, we are applying the separation of variables to tackle the equation for a cylindrical conductor with a specific set of boundary conditions.

To progress, we substitute the separated variables back into Laplace's equation and adjust for constants, which leads to the emergence of the terms \(X'' = k^2X\) and \(Y'' = -k^2Y\), indicating harmonic functions. Such functions are well understood and have solutions in the form of sine, cosine, hyperbolic sine, and hyperbolic cosine functions. We then fit these solutions into the boundary conditions as a next step. The constants that we derive from these fittings make the solutions conform to the physical reality of the conductor's electrostatic scenario.

Cylindrical Conductor Potential

After implementing the concept of separation of variables and applying appropriate boundary conditions, we attain a series representation for the potential inside the cylindrical conductor as \(V(x,y) = \sum_{n=1}^\infty {\frac{2V_{0}}{\sinh(n\pi)} \sin\left(\frac{n\pi y}{a}\right) \sinh\left(\frac{n\pi x}{a}\right)}\). This infinite series expresses the sum of possibilities for the potential at any point \((x, y)\) inside the conductor.

Each term of the series corresponds to a 'mode' of the potential, and the complete solution is a superposition of these modes, weighted by the boundary conditions. Through observation, we understand that because of the conductive material, the potential functions as a sum of sinusoids along the y-axis and hyperbolic sinusoids along the x-axis. These functions reflect the behavior of the electric potential across the square cross-sections and through the depth of the conductor, providing insight into electrostatic distribution and the influence of geometry and boundary conditions on it.