Question

Let \(V=(\cos 2 \phi) / \rho\) in free space. (a) Find the volume charge density at point \(A\left(0.5,60^{\circ}, 1\right) .(b)\) Find the surface charge density on a conductor surface passing through the point \(B\left(2,30^{\circ}, 1\right)\).

Short answer

Based on the given electric potential function \(V(\rho, \phi) = \frac{\cos 2\phi}{\rho}\) in polar coordinates, we computed the volume charge density at point A as \(\rho_{v}(A) = -20\epsilon_0\) and the surface charge density at point B as \(\sigma_B = -\frac{1}{2}\epsilon_0\).

Step-by-step solution

Compute the Laplacian of the potential function in polar coordinates

First, we need to find the expression of the Laplacian of the potential function in polar coordinates. The general formula for the Laplacian in polar coordinates is given by: $$\nabla^2 V(\rho, \phi) = \frac{1}{\rho}\frac{\partial}{\partial \rho}(\rho \frac{\partial V}{\partial \rho}) + \frac{1}{\rho^2} \frac{\partial^2 V}{\partial \phi^2}$$ Now, let's compute all the derivatives.

Compute the required derivatives

We need to compute the derivatives of the potential function \(V\) as follows: First derivative with respect to \(\rho\): $$ \frac{\partial V}{\partial \rho} = -\frac{\cos 2\phi}{\rho^2} $$ First derivative with respect to \(\phi\): $$ \frac{\partial V}{\partial \phi} = -\frac{2\sin 2\phi}{\rho} $$ Second derivative with respect to \(\phi\): $$ \frac{\partial^2 V}{\partial \phi^2} = -\frac{4\cos 2\phi}{\rho} $$ Now, let's compute the Laplacian.

Plug the derivatives into the Laplacian formula

Using the computed derivatives, we can evaluate the Laplacian formula as follows: $$\nabla^2 V(\rho, \phi) = \frac{1}{\rho}\frac{\partial}{\partial \rho}(\rho \cdot -\frac{\cos 2\phi}{\rho^2}) + \frac{1}{\rho^2} \cdot -\frac{4\cos 2\phi}{\rho}$$ Simplifying this expression, we get: $$\nabla^2 V(\rho, \phi) = -\frac{\cos 2\phi}{\rho^3} - \frac{4\cos 2\phi}{\rho^3} = -\frac{5\cos 2\phi}{\rho^3}$$

Compute the volume charge density at point A

To find the volume charge density at point A, we need to evaluate the Laplacian at the given coordinates A \((0.5, 60^\circ, 1)\). We have: $$\rho = 0.5, \phi = 60^\circ = \frac{\pi}{3}$$ $$\nabla^2 V(\rho, \phi) = -\frac{5\cos(2\phi)}{\rho^3} = -\frac{5\cos\left(2 \cdot \frac{\pi}{3}\right)}{\left(0.5\right)^3} = -\frac{5}{\left(0.5\right)^3}$$ The volume charge density at point A equals to: $$\rho_{v}(A) = \epsilon_0 \nabla^2 V(\rho, \phi) = -20\epsilon_0$$

Compute the gradient of the potential function

Next, we need to compute the gradient \(\nabla V\) of the potential function: $$\nabla V = \frac{\partial V}{\partial \rho}\hat{\rho} + \frac{1}{\rho}\frac{\partial V}{\partial \phi}\hat{\phi}$$ Substituting the values of the derivatives, we get: $$\nabla V = -\frac{\cos 2\phi}{\rho^2}\hat{\rho} -\frac{2\sin 2\phi}{\rho}\hat{\phi}$$

Evaluate the gradient at point B

Now, we need to evaluate the gradient at point B \((2, 30^\circ, 1)\). We have: $$\rho = 2, \phi = 30^\circ = \frac{\pi}{6}$$ So, the gradient at point B is: $$\nabla V(B) = -\frac{\cos\left(2\cdot \frac{\pi}{6}\right)}{(2)^2}\hat{\rho} -\frac{2\sin\left(2\cdot \frac{\pi}{6}\right)}{2}\hat{\phi} = -\frac{1}{4}\hat{\rho} -\frac{1}{2}\hat{\phi}$$

Find the surface charge density at point B

To find the surface charge density at point B, let's consider the normal component of the gradient of the potential function, since the field lines start from the positive charge and end on the negative charge. Therefore, the surface charge density (sigma) at point B is given by: $$\sigma_B = \epsilon_0 \nabla V(B)_n$$ For simplicity purposes, it is assumed that the surface of the conductor is not curved, so we only need to take the component of the gradient along the \(\hat{\phi}\) direction as the normal component. $$\sigma_B = \epsilon_0 \left(-\frac{1}{2}\right) = -\frac{1}{2}\epsilon_0$$ So, the surface charge density at point B is \(-\frac{1}{2}\epsilon_0\).

Key concepts

Laplacian in Polar Coordinates

Understanding the Laplacian in polar coordinates is essential when dealing with problems in electromagnetism and other areas of physics where cylindrical symmetry is present. The Laplacian operator, denoted as \(abla^2\), is crucial because it helps us express the second-order differential of functions, such as electrostatic potential in a curved coordinate system like polar coordinates.

The general form of the Laplacian in polar coordinates is expressed by the formula:
\[abla^2 V(\rho, \rho) = \frac{1}{\rho}\frac{\partial}{\partial \rho}(\rho \frac{\partial V}{\partial \rho}) + \frac{1}{\rho^2} \frac{\partial^2 V}{\partial \phi^2}\]
Through this formula, the Laplacian considers both radial variation (with \(\rho\)) and angular variation (with \(\rho\)) of the potential function \(V\). When solving for electrostatic scenarios, as shown in the original exercise, computing the Laplacian in polar coordinates allows us to determine how charge is distributed in space.

Electrostatic Potential

Electrostatic potential, often denoted as \(V\), is a scalar quantity that represents the electric potential energy per unit charge at a certain point in space. It's a fundamental concept in electrostatics, providing insight into how electric fields influence charged particles. The electrostatic potential due to a charge distribution is determined by the integral of the electric field along a path.

In the given exercise, we see a potential function defined as \(V=(\cos 2 \phi) / \phi\). This particular form tells us how potential varies with distance (\(\rho\)) and angle (\(\rho\)). To understand the charge distribution that leads to this potential, we take the Laplacian; a non-zero Laplacian implies the presence of charge, while a zero Laplacian (in free space) would imply regions without charge. Thus, by analyzing the potential function and its derivatives, we can infer characteristics about the source of an electric field.

Surface Charge Density

Surface charge density, denoted as \(\sigma\), quantifies the amount of electric charge per unit area on a surface. This concept relates closely to boundary conditions in electrostatics, particularly when dealing with conductors and dielectric surfaces. The surface charge density is critical when transitioning from one medium to another, as it can influence both the electric field and electrostatic potential.

As demonstrated in the solution steps for point B, we calculated the gradient of the potential to find the electric field. In the context of conductors, the surface charge density is related to the electric field just outside the surface. The component of the electric field normal to the surface gives rise to the surface charge density. By using the gradient of the potential and getting its normal component, we can compute the surface charge density on the conductor at that point, which is crucial for determining the behavior of electric fields around the conductor's surface.