STEM Problem Detail

Problem Statement

Prompt

Provide me Step By Step Solution for Problem 3.74

Solution Rendering...

**3.74 Solution**

A point charge $$q$$ is located at the centre of a ball made of a uniform isotropic
dielectric with permittivity $$\varepsilon$$.

**Electric field inside the dielectric**

By symmetry, the electric field is radial.
Using Gauss’s law for the electric displacement,

$$
\oint \mathbf{D}\cdot d\mathbf{S} = q.
$$

For a spherical Gaussian surface of radius $$r$$,

$$
D \cdot 4\pi r^2 = q,
$$

hence

$$
\mathbf{D} = \frac{q}{4\pi r^2}\,\mathbf{r}.
$$

The electric field is related to $$\mathbf{D}$$ by

$$
\mathbf{E} = \frac{\mathbf{D}}{\varepsilon}.
$$

Thus,

$$
\mathbf{E} = \frac{q}{4\pi\varepsilon r^2}\,\mathbf{r}.
$$

**Polarization vector**

The polarization is defined as

$$
\mathbf{P} = \mathbf{D} - \mathbf{E}.
$$

Substituting $$\mathbf{D}$$ and $$\mathbf{E}$$,

$$
\mathbf{P}
= \frac{q}{4\pi r^2}\mathbf{r}
- \frac{q}{4\pi\varepsilon r^2}\mathbf{r}.
$$

Factoring,

$$
\mathbf{P}
= \frac{\varepsilon-1}{4\pi\varepsilon}\,
\frac{q}{r^2}\,\mathbf{r}.
$$

**Bound charge inside a sphere of radius $$r$$**

The bound volume charge density is

$$
\rho_b = -\nabla\cdot\mathbf{P}.
$$

The total bound charge inside radius $$r$$ is

$$
q' = \int \rho_b\,dV
= -\oint \mathbf{P}\cdot d\mathbf{S}.
$$

Using spherical symmetry,

$$
q' = -P \cdot 4\pi r^2.
$$

Substituting $$\mathbf{P}$$,

$$
q' = -4\pi r^2
\cdot
\frac{\varepsilon-1}{4\pi\varepsilon}
\frac{q}{r^2}.
$$

Simplifying,

$$
q' = -\frac{\varepsilon-1}{\varepsilon}\,q.
$$

Loading solution...

Problem #26