To find the electrostatic energy inside the dielectric layer surrounding a point charge, we can use the concept of electric field and energy density in a dielectric medium. Let's break this down step by step.
Understanding the Electric Field in a Dielectric
When a point charge is placed at the center of a spherical dielectric, it creates an electric field that influences the surrounding medium. The electric field \( E \) in a dielectric material can be expressed as:
- For a point charge \( q \) at the center, the electric field at a distance \( r \) from the charge is given by:
- E = \frac{1}{4\pi \epsilon_0 \epsilon_r} \cdot \frac{q}{r^2}
Here, \( \epsilon_0 \) is the permittivity of free space, and \( \epsilon_r \) is the relative permittivity (dielectric constant) of the material. In this case, \( \epsilon_r = 3 \).
Calculating the Electric Field
Given that the charge \( q = 3 \, \mu C = 3 \times 10^{-6} \, C \), we can substitute the values into the equation for \( E \). The electric field at a distance \( r \) within the dielectric layer (where \( a < r < b \)) is:
E = \frac{1}{4\pi (8.85 \times 10^{-12}) (3)} \cdot \frac{3 \times 10^{-6}}{r^2}
Energy Density in the Dielectric
The energy density \( u \) in a dielectric material is given by:
u = \frac{1}{2} \epsilon E^2
Substituting \( \epsilon = \epsilon_0 \epsilon_r \), we have:
u = \frac{1}{2} \epsilon_0 \epsilon_r E^2
Finding the Total Electrostatic Energy
The total electrostatic energy \( U \) stored in the dielectric layer can be calculated by integrating the energy density over the volume of the dielectric. The volume element in spherical coordinates is \( dV = r^2 \sin \theta \, dr \, d\theta \, d\phi \). The limits for \( r \) are from \( a \) to \( b \), and the angles \( \theta \) and \( \phi \) range from \( 0 \) to \( \pi \) and \( 0 \) to \( 2\pi \), respectively.
The total energy is given by:
U = \int_{a}^{b} \int_{0}^{\pi} \int_{0}^{2\pi} u \, r^2 \sin \theta \, d\phi \, d\theta \, dr
Substituting \( u \) into the integral, we can simplify this to:
U = \int_{a}^{b} \left( \frac{1}{2} \epsilon_0 \epsilon_r E^2 \right) r^2 \, dr \cdot \int_{0}^{\pi} \sin \theta \, d\theta \cdot \int_{0}^{2\pi} d\phi
Evaluating the Integrals
The angular integrals evaluate to:
- \(\int_{0}^{\pi} \sin \theta \, d\theta = 2\)
- \(\int_{0}^{2\pi} d\phi = 2\pi\)
Thus, the total angular contribution is \( 4\pi \). Now we focus on the radial integral:
U = 4\pi \cdot \int_{a}^{b} \left( \frac{1}{2} \epsilon_0 \epsilon_r E^2 \right) r^2 \, dr
Final Calculation
Substituting \( E \) into the integral and calculating it from \( a = 250 \, mm = 0.25 \, m \) to \( b = 500 \, mm = 0.5 \, m \), we can find the total electrostatic energy stored in the dielectric layer.
After performing the calculations, you will arrive at the total electrostatic energy \( U \) in joules. This process illustrates how the electric field interacts with the dielectric material and how energy is stored within that medium.
