Question

A charge $q$ is uniformly distributed over the volume of a dielectric sphere of radius $a$. If the dielectric constant $ \epsilon _r= 2$, then the ratio of the electrostatic energy stored inside the sphere to that stored outside is _________ . (Round off to I decimal place).

Answer 1

Correct Answer: 0.1

1. Concept Overview

The electrostatic energy ($U$) stored in a region with an electric field $\mathbf{E}$ is given by the integral over the volume:

$$U = \int \frac{1}{2} \epsilon E^2 \, dV$$

where:

$\epsilon = \epsilon_0 \epsilon_r$ is the permittivity of the medium.

$dV = 4\pi r^2 dr$ is the differential volume element for spherical symmetry.

We need to calculate the energy stored inside the sphere ($U_{in}$) and outside the sphere ($U_{out}$) separately and then find the ratio $U_{in}/U_{out}$.

2. Electric Field Calculations

Case A: Outside the sphere ($r > a$)

For points outside the sphere, the electric field behaves as if all charge is concentrated at the center. The medium outside is usually vacuum/air ($\epsilon_r = 1$).

$$E_{out} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}$$

Case B: Inside the sphere ($r < a$)

Using Gauss's Law for a dielectric medium:

$$\oint \mathbf{D} \cdot d\mathbf{A} = q_{enclosed}$$

Where $\mathbf{D} = \epsilon \mathbf{E} = \epsilon_0 \epsilon_r \mathbf{E}$.

The charge enclosed within radius $r$ is a fraction of the total charge $q$ based on volume ratio:

$$q_{enclosed} = q \left( \frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi a^3} \right) = q \frac{r^3}{a^3}$$

Applying Gauss's Law:

$$\epsilon_0 \epsilon_r E_{in} (4\pi r^2) = q \frac{r^3}{a^3}$$

$$E_{in} = \frac{1}{4\pi\epsilon_0 \epsilon_r} \frac{q r}{a^3}$$

3. Calculate Energy Stored Outside ($U_{out}$)

Limits of integration: $r = a$ to $r = \infty$.

Permittivity outside: $\epsilon = \epsilon_0$.

$$U_{out} = \int_{a}^{\infty} \frac{1}{2} \epsilon_0 E_{out}^2 (4\pi r^2) \, dr$$

Substitute $E_{out} = \frac{q}{4\pi\epsilon_0 r^2}$:

$$U_{out} = \frac{1}{2} \epsilon_0 \int_{a}^{\infty} \left( \frac{q}{4\pi\epsilon_0 r^2} \right)^2 4\pi r^2 \, dr$$

$$U_{out} = \frac{1}{2} \epsilon_0 \frac{q^2}{16\pi^2 \epsilon_0^2} 4\pi \int_{a}^{\infty} \frac{1}{r^2} \, dr$$

$$U_{out} = \frac{q^2}{8\pi \epsilon_0} \left[ -\frac{1}{r} \right]_a^{\infty}$$

$$U_{out} = \frac{q^2}{8\pi \epsilon_0} \left( 0 - \left(-\frac{1}{a}\right) \right)$$

$$U_{out} = \frac{q^2}{8\pi \epsilon_0 a}$$

4. Calculate Energy Stored Inside ($U_{in}$)

Limits of integration: $r = 0$ to $r = a$.

Permittivity inside: $\epsilon = \epsilon_0 \epsilon_r$.

$$U_{in} = \int_{0}^{a} \frac{1}{2} (\epsilon_0 \epsilon_r) E_{in}^2 (4\pi r^2) \, dr$$

Substitute $E_{in} = \frac{q r}{4\pi\epsilon_0 \epsilon_r a^3}$:

$$U_{in} = \frac{1}{2} (\epsilon_0 \epsilon_r) \int_{0}^{a} \left( \frac{q r}{4\pi\epsilon_0 \epsilon_r a^3} \right)^2 4\pi r^2 \, dr$$

$$U_{in} = \frac{1}{2} (\epsilon_0 \epsilon_r) \frac{q^2}{(16\pi^2 \epsilon_0^2 \epsilon_r^2 a^6)} 4\pi \int_{0}^{a} r^4 \, dr$$

Simplify constants:

$$U_{in} = \frac{q^2}{8\pi \epsilon_0 \epsilon_r a^6} \int_{0}^{a} r^4 \, dr$$

$$U_{in} = \frac{q^2}{8\pi \epsilon_0 \epsilon_r a^6} \left[ \frac{r^5}{5} \right]_0^a$$

$$U_{in} = \frac{q^2}{8\pi \epsilon_0 \epsilon_r a^6} \cdot \frac{a^5}{5}$$

$$U_{in} = \frac{q^2}{40\pi \epsilon_0 \epsilon_r a}$$

5. Calculate the Ratio

We need the ratio $\frac{U_{in}}{U_{out}}$.

$$\text{Ratio} = \frac{ \frac{q^2}{40\pi \epsilon_0 \epsilon_r a} }{ \frac{q^2}{8\pi \epsilon_0 a} }$$

Canceling common terms ($q^2, \pi, \epsilon_0, a$):

$$\text{Ratio} = \frac{ \frac{1}{40 \epsilon_r} }{ \frac{1}{8} }$$

$$\text{Ratio} = \frac{8}{40 \epsilon_r} = \frac{1}{5 \epsilon_r}$$

Given $\epsilon_r = 2$:

$$\text{Ratio} = \frac{1}{5(2)} = \frac{1}{10}$$

$$\text{Ratio} = 0.1$$

Final Answer

The ratio of the electrostatic energy stored inside the sphere to that stored outside is 0.1.