Chapter 4: Problem 10
A sphere of radius \(R\) carries a polarization $$ \mathbf{P}(\mathbf{r})=k \mathbf{r} $$ where \(k\) is a constant and \(\mathbf{r}\) is the vector from the center. (a) Calculate the bound charges \(\sigma_{b}\) and \(\rho_{b}\). (b) Find the field inside and outside the sphere.
Short Answer
Expert verified
\(\rho_b = -3k\); \(\sigma_b = kR\); \(\mathbf{E}_{\text{inside}} = 0\); \(\mathbf{E}_{\text{outside}}\) follows dipole field equation.
Step by step solution
01
Understand Polarization
The polarization given is \( \mathbf{P}(\mathbf{r}) = k \mathbf{r} \), where \( k \) is a constant and \( \mathbf{r} \) represents the position vector from the center of the sphere. Polarization indicates the density of electric dipole moments within the material.
02
Calculate Bound Volume Charge Density \( \rho_b \)
The bound volume charge density can be calculated using \( \rho_{b} = -abla \cdot \mathbf{P} \). For the given polarization \( \mathbf{P}(\mathbf{r}) = k \mathbf{r} \), \( abla \cdot \mathbf{P} = abla \cdot (k \mathbf{r}) = k abla \cdot \mathbf{r} = k \, (3) = 3k \),so \( \rho_{b} = -3k \).
03
Calculate Bound Surface Charge Density \( \sigma_b \)
The bound surface charge density is given by \( \sigma_b = \mathbf{P} \cdot \hat{n} \), where \( \hat{n} \) is the outward unit normal to the surface.At the surface of the sphere, \( \mathbf{r} = R \hat{n} \) (where \( R \) is the radius and \( \hat{n} \) is the unit normal). Therefore, \( \mathbf{P}(R \hat{n}) = k R \hat{n} \),thus \( \sigma_{b} = (k R \hat{n}) \cdot \hat{n} = k R \).
04
Find Electric Field Inside the Sphere
Inside the sphere, by symmetry and the solution of Laplace's equation with the given boundary conditions, the electric field \( \mathbf{E}_{\text{inside}} \) is zero. The polarization causes no electric field within a uniformly polarized sphere.
05
Find Electric Field Outside the Sphere
Outside the sphere, the sphere can be treated as a point dipole at the origin with dipole moment \( \mathbf{p} = \frac{4\pi}{3} R^3 \mathbf{P} = \frac{4\pi}{3} R^3 k \mathbf{r} \). The electric field due to a dipole at a distance \( r \) from the center in the radial direction is given by:\[ \mathbf{E}_{\text{outside}} = \frac{1}{4\pi\varepsilon_0} \left[ \frac{3(\mathbf{p} \cdot \mathbf{r}) \mathbf{r}}{r^5} - \frac{\mathbf{p}}{r^3} \right] \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Bound charges
Bound charges arise in a dielectric material due to polarization. When you have a polarized object, like a sphere, this means the bound charges are not free to move. They merely shift slightly to create an opposing electric field. Bound volume charge density, represented by \( \rho_b \), is related to the divergence of the polarization \( \mathbf{P} \). In mathematical terms, we calculate it using \( \rho_{b} = -abla \cdot \mathbf{P} \). For the polarization \( \mathbf{P}(\mathbf{r}) = k \mathbf{r} \), the divergence \( abla \cdot \mathbf{r} \) is simply \( 3 \), so the volume bound charge density becomes \( \rho_b = -3k \). This means the volume within the sphere carries a density of bound charges equal to \( -3k \).Bound surface charge density \( \sigma_b \) occurs at the surface of the dielectric and is calculated using \( \sigma_b = \mathbf{P} \cdot \hat{n} \). Here, \( \hat{n} \) is the outward normal vector on the sphere's surface. With \( \mathbf{P}(R \hat{n}) = k R \hat{n} \) at the surface, the bound surface charge density simplifies to \( \sigma_b = kR \). Thus, the surface charge density is directly proportional to the sphere's radius and the constant \( k \). This helps explain how the charges align themselves at the boundary of the polarized sphere.
Electric field inside and outside a sphere
Understanding the electric field both inside and outside a polarized sphere is key to predicting how electromagnetically charged objects interact. **Electric Field Inside the Sphere**Inside the polarized sphere, symmetry considerations simplify the problem. Due to uniform polarization, the internal electric field, \( \mathbf{E}_{\text{inside}} \), is zero. This means that the polarization field cancels out any electric field within the sphere. It points to a fundamental concept - within a uniformly polarized sphere, charges adjust perfectly to neutralize internal fields.**Electric Field Outside the Sphere**Things change when considering the field outside the sphere. Here, the whole sphere behaves like a point dipole located at its center. The strength and direction of this dipole is determined by the dipole moment \( \mathbf{p} = \frac{4\pi}{3} R^3 \mathbf{P} \). Using the formula for the electric field from a dipole at a point distance \( r \), we calculate:\[ \mathbf{E}_{\text{outside}} = \frac{1}{4\pi\varepsilon_0} \left[ \frac{3(\mathbf{p} \cdot \mathbf{r}) \mathbf{r}}{r^5} - \frac{\mathbf{p}}{r^3} \right] \]This formula encapsulates the decay and behavior of the field at distances far from the sphere, behaving similarly to how a point charge would behave but tailored to a dipole.
Electric dipole moment
An electric dipole moment is fundamental in describing the separation of charge in system, particularly in polarized materials. It measures how a pair of charges interacts with electric fields and is a vector pointing from the negative to the positive charge.In a continuous body such as our sphere, the electric dipole moment \( \mathbf{p} \) can be found by integrating over the polarized volume. For simplicity, our sphere's electrostatic problem allows us to use the formula \( \mathbf{p} = \frac{4\pi}{3} R^3 \mathbf{P} \), which effectively gives:\[ \mathbf{p} = \frac{4\pi}{3} R^3 k \mathbf{r} \]This equation tells us how the dipole's strength increases with both the radius of the sphere and the polarization.Key characteristics of a dipole:
- The dipole moment vector defines the direction of the resultant electric field.
- The magnitude tells us the strength of the field at any point.
- A dipole's influence diminishes quickly with distance, as the field typically varies with the cube of the distance.