Question

Find a complex potential in the \(z\)-plane appropriate to a physical situation in which the half-plane \(x>0, y=0\) has zero potential and the half-plane \(x<0\) \(y=0\) has potential \(V .\) By making the transformation \(w=a\left(z+z^{-1}\right) / 2\), with \(a\) real and positive, find the electrostatic potential associated with the half-plane \(r>a, s=0\) and the half-plane \(r<-a, s=0\) at potentials 0 and \(V\) respectively.

Short answer

The potential is associated with \(r > a\) (potential 0) and \(r < -a\) (potential \(V\)) at \(s = 0\) in the \(w\)-plane.

Step-by-step solution

Understand the given transformation

The transformation given in the problem is \(w = \frac{a}{2}(z + z^{-1})\). Here, \(z\) represents a point in the complex plane, and \(w\) is a transformation of \(z\). We need to apply this transformation properly to understand how the potential fields in the \(z\)-plane map to the \(w\)-plane.

Apply the transformation to the \(z\)-plane

Given that \(z = x + iy\) (where \(i\) is the imaginary unit), substitute \(z\) into the transformation \(w = \frac{a}{2}(z + z^{-1})\).Hence, \(w = \frac{a}{2}\bigg((x+iy) + \frac{1}{x+iy}\bigg)\).

Simplify the expression for \(w\)

Rewrite the inverse of \(z\): \(z^{-1} = \frac{1}{x + iy} = \frac{x-iy}{x^2+y^2}.\) Now add the expressions together: \(w = \frac{a}{2}\bigg(x + iy + \frac{x-iy}{x^2+y^2}\bigg)\). Simplify further to separate the real and imaginary parts.

Consider the boundaries of the half-planes

For \(x > 0\), \(y = 0\) corresponds to points on the positive real axis, so the transformation \(w = \frac{a}{2}(x + \frac{1}{x})\) must result in zero potential in the \(w\)-plane. For \(x < 0\), \(y = 0\), the transformation should yield a potential of \(V\) in the \(w\)-plane.

Map the zero potential and potential \(V\)

For \(x > 0\), \(w = \frac{a}{2}(x + \frac{1}{x})\) for \(y = 0\). This implies that the real axis in the \(z\)-plane for positive \(x\) transforms to a line in the \(w\)-plane with zero potential. For \(x < 0\), \(w = \frac{a}{2}(x + \frac{1}{x})\) for \(y = 0\) but \(x\) is negative. This implies that the real axis in the \(z\)-plane for negative \(x\) transforms to a line in the \(w\)-plane with potential \(V\).

Verify the potential in \(w\)-plane

The transformation \(w = \frac{a}{2}(z + z^{-1})\) implies that in the \(w\)-plane, the potential is 0 for \(r > a, s = 0\) (real axis, \(r\) positive) and \(V\) for \(r < -a, s = 0\) (real axis, negative \(r\)). This matches the given condition.

Key concepts

conformal mapping

A conformal mapping is a transformation that preserves angles and the shape of infinitesimally small figures in the complex plane. This transformation helps in visualizing complex potentials and fields in electrostatics by simplifying the geometric configuration. By transforming the coordinates, it becomes easier to analyze certain physical situations such as the boundaries of electrical potentials. In the given problem, the transformation used is given by:

\[ w = \frac{a}{2}(z + z^{-1}) \]

Here, the variable \(z\) denotes a point in the complex plane, and \(w\) is the image of \(z\) under this conformal map. Conformal mappings are powerful because they can transform complex boundary conditions into simpler forms while retaining essential properties.

electrostatics

Electrostatics deals with the study of electric charges at rest and the electric fields and potentials they produce. In the context of the problem, we consider a situation where different regions of a plane have distinct electric potentials. Particularly, the half-plane where \(x > 0\) (positive real axis) has zero potential, while the half-plane where \(x < 0\) (negative real axis) has a potential of \(V\).

Understanding how charges and potentials distribute themselves around boundaries helps to identify the electric potential distribution in more complex geometries.

In practice, we use conformal mapping to transform the initial complex boundary set-up in a way that makes the mathematics simpler while ensuring the physical situation's consistency.

half-plane potential

In this problem, we deal with a half-plane potential, where different half-planes have different potentials. In electrostatics, potential is a scalar quantity representing the electric potential energy per unit charge at a point in space.

Specifically, the problem involves:

• The half-plane \(x > 0, y = 0\) with zero potential.
• The half-plane \(x < 0, y = 0\) with a potential \(V\).

To solve this, we use the conformal mapping to transform the coordinates in the complex plane, simplifying how we handle boundary conditions for electrostatic potentials. By applying the given transformation, the regions of different potentials in the original plane (z-plane) are transformed to simpler regions in the mapped plane (w-plane), while maintaining the physical conditions given in the problem.