Question

Given the potential field \(V=\left(A \rho^{4}+B \rho^{-4}\right) \sin 4 \phi:(a)\) Show that \(\nabla^{2} V=0\). (b) Select \(A\) and \(B\) so that \(V=100 \mathrm{~V}\) and \(|\mathbf{E}|=500 \mathrm{~V} / \mathrm{m}\) at \(P(\rho=1\), \(\left.\phi=22.5^{\circ}, z=2\right) .\)

Short answer

Upon solving the system of equations, we find two possible sets of values for A and B. They are: Set 1: A = 27.5737 B = 6.2055 Set 2: A = 6.2055 B = 27.5737 In conclusion, we have shown that the Laplacian of the given potential field is zero, meaning it satisfies Laplace's equation. Moreover, we determined two possible sets of values for A and B that satisfy the given conditions for potential V and electric field E at point P. Both sets of values are valid which indicates that there are multiple solutions for A and B in this case.

Step-by-step solution

(a) Finding the Laplacian of V

In order to prove that \(\nabla^2 V = 0\), we have to find the Laplacian of the given potential. In cylindrical coordinates, the Laplacian of a scalar function V is given by $$\nabla^2 V = \frac{1}{\rho}\frac{\partial}{\partial \rho}\left( \rho \frac{\partial V}{\partial \rho} \right)+\frac{1}{\rho^2}\frac{\partial^2 V}{\partial \phi^2} + \frac{\partial^2 V}{\partial z^2}.$$ Now, let's calculate each term separately.

Calculating \( \rho \frac{\partial V}{\partial \rho} \)

$$\rho \frac{\partial V}{\partial \rho} = \rho \frac{\partial}{\partial \rho} ( (A \rho^4 + B \rho^{-4})\sin(4\phi)) = ( 4A\rho^4 - 4B\rho^{-4})\sin(4\phi).$$

Calculating \(\frac{1}{\rho}\frac{\partial}{\partial \rho}\left( \rho \frac{\partial V}{\partial \rho} \right)\)

$$\frac{1}{\rho}\frac{\partial}{\partial \rho}\left( \rho \frac{\partial V}{\partial \rho} \right) = \frac{\partial}{\partial \rho}( 4A\rho^3 - 4B\rho^{-5})\sin(4\phi) = ( 12A\rho^2 + 20B\rho^{-6})\sin(4\phi).$$

Calculating \(\frac{1}{\rho^2}\frac{\partial^2 V}{\partial \phi^2} \)

$$\frac{1}{\rho^2} \frac{\partial^2 V}{\partial \phi^2} = \frac{1}{\rho^2} \frac{\partial^2}{\partial \phi^2} ( (A \rho^4 + B \rho^{-4})\sin(4\phi)) = -16 (A \rho^2 + B \rho^{-6})\sin(4\phi).$$

Laplacian V

Now let's sum up all the terms: $$\nabla^2 V = (12A\rho^2 + 20B\rho^{-6})\sin(4\phi) -16 (A \rho^2 + B \rho^{-6})\sin(4\phi) =0.$$ Thus, we showed that \(\nabla^2 V = 0\).

(b) Selecting A and B

To find the values of A and B, we have to use the given conditions at the point P. 1) V(P) = 100 V: Substitute \(\rho = 1\) and \(\phi = 22.5^\circ\) in the potential field equation: $$100 = (A + B)\sin(4(22.5^\circ)).$$ 2) \(|E|\) = 500 V/m at P: The electric field is related to the potential by \(E_\rho = -\frac{\partial V}{\partial \rho}\), so we need to find \(E_\rho\) at P. It is important to note that, because of the symmetry of the problem and the given potential equation, \(E_\phi\) will be zero. $$E_\rho = -\frac{\partial V}{\partial \rho}=- 4(A-B) \sin(4\phi).$$ Substitute \(\rho = 1\) and \(\phi = 22.5^\circ\) in \(E_\rho\) equation: $$|E_\rho|=500=4|A - B|\sin(4(22.5^\circ)).$$ Now, we have a system of 2 equations with 2 unknowns, A and B: $$ (A + B)\sin(4(22.5^\circ)) = 100$$ $$4|(A - B)\sin(4(22.5^\circ))| = 500$$ Solving this system of equations, we can find the values of constants A and B. It is important to note that there can be multiple solutions for A and B, as the absolute value is involved in the second equation.

Key concepts

Electromagnetic Potential: Understanding the Foundation of Electric Fields

Electromagnetic potential represents the potential energy per unit charge at a point in space due to an electric field. In electromagnetics, it's often denoted as 'V' and is a fundamental concept linking the electric field and potential energy.

Consider an electric field. When a charge moves within this field, it experiences a force, and consequently, either gains or loses potential energy depending on the direction of movement. The concept of electromagnetic potential helps us calculate the energy without knowing the specific path taken by the charge, which is why it's so powerful. In the provided exercise, the potential field is given as a function of cylindrical coordinates \( \rho \) and \( \phi \), showcasing its variation with distance and angle from the origin.

One critical aspect of electromagnetic potential is its relation to the electric field intensity \( \vec{E} \). This vector field describes the force per unit charge at any given point in space, and can be determined by the negative gradient of the electromagnetic potential, which leads us to the expression \( \vec{E} = -abla V \). In the exercise, the determination of electromagnetic potential is essential for subsequent calculations of electrical field intensity.

Electrical Field Intensity: Quantifying the Strength of Electric Fields

Electrical field intensity, denoted by \( \vec{E} \), is the force experienced by a unit positive charge in an electric field. It's a vector quantity, meaning it has both magnitude and direction, and is a fundamental concept in understanding how charges interact with fields.

In the provided exercise, the electrical field intensity is explored through the manipulation of the electromagnetic potential. By taking the negative gradient of the potential \( V \), we expose the relationship between these quantities. Specifically, for a radial function in a cylindrical coordinate system, the radial component of the electric field \( E_\rho \) can be found using \( E_\rho = -\frac{\partial V}{\partial \rho} \).

The question asks for the electric field intensity at a particular point \( P \). By substituting the values of \( \rho \) and \( \phi \) into the derivative of the potential function, one can find the intensity at that point. The intensity is also related to the potential constants \( A \) and \( B \) from the exercise, showcasing that the potential shape affects the strength and direction of the electric field.

Cylindrical Coordinate System: A 3D Framework for Electromagnetics

The cylindrical coordinate system is a three-dimensional system that expands upon the idea of polar coordinates by adding an extra dimension for height (z). It's particularly useful in electromagnetics for describing fields and potentials with cylindrical symmetry.

In this system, any point in space is represented by three values: \( \rho \), which is the radial distance from the central axis; \( \phi \) the angle around the axis starting from a fixed direction; and \( z \), the height above the base plane. This framework offers a convenient way to express the Laplacian of a scalar function, which is essential for solving electromagnetic problems.

For finding the Laplacian in the cylindrical coordinate system, you must consider the contributions from all directions - radial, angular, and axial. The exercise showcases this concept, where the Laplacian of \( V \) must demonstrate that it equals zero for all points in space. This approach is particularly helpful when dealing with cylindrical objects like cables or wires, where symmetry around the axis simplifies equations and calculations, allowing for easier analysis of the electric and magnetic fields involved.