PHYS212 : Lectures

Subsections

• 5.4.1 Energy Density
• 5.4.2 Intensity and the Poynting Vector
• 5.4.3 Radiation Pressure

5.4 Energy, Power, and Intensity

5.4.1 Energy Density

• The energy density - the energy per unit volume - in an electric field can be obtained by considering the field of an ideal parallel plate capacitor.

• The energy stored in a charged capacitor is equivalent to the work done in loading the charges onto the plates.
  

  $$U = \int_0^V q(V) dV = \int_0^V CV dV = \frac{1}{2}CV^2$$ (5.19)

  
  

• The capacitance of a parallel plate capacitor is given by^5.2
  

  $$C = \epsilon_o \frac{A}{d}$$ (5.20)

  
  
  where $A$ is the area of the plates and $d$ is the plate separation.

• Eq.'s 5.19 and 5.20 combine to give the energy density of the electric field
  
  $$w_E = \frac{U_E}{Ad} = \frac{1}{2} \epsilon_o \frac{A}{Ad^2} V^2$$
  
  
  which can be written in terms of the electric field $E=\frac{V}{d}$ as
  

  $$w_E = \frac{1}{2} \epsilon_o E^2$$ (5.21)

  
  

• Similarly, the energy density in a magnetic field can be obtained by considering the field of a long solenoid.

• The energy stored in an inductor is equivalent to the work done in bringing the current it carries up from zero. The power delivered to an inductor
  
  $$\frac{dU}{dt} = -Vi$$
  
  

• Using Faraday's law, we have
  
  $$\frac{dU}{dt} = -iL\frac{di}{dt}$$
  
  
  which we can integrate over time to find
  

  $$U = L \int_0^I i di = \frac{1}{2} LI^2$$ (5.22)

  
  

• The inductance of a solenoid is given by^5.3
  

  $$L = \frac{\mu_o N^2 A}{\ell}$$ (5.23)

  
  
  where $N$ is the number of turns, and $\ell$ is the length of the solenoid.

• Eq.'s 5.22 and 5.23 combine to give the energy density of the magnetic field
  
  $$w_B = \frac{U_B}{A\ell} = \frac{1}{2} \frac{\mu_o N^2 I^2}{\ell^2}$$
  
  
  which can be written in terms of the magnetic field of a solenoid
  
  $$B = \frac{\mu_o N i}{\ell}$$
  
  
  as
  

  $$w_B = \frac{B^2}{2\mu_o}$$ (5.24)

  
  

• Eq.'s 5.21 and 5.24 combine to give the energy density of an electromagnetic wave,
  

  $$w = \frac{1}{2} \epsilon_o E^2 + \frac{B^2}{2\mu_o}$$ (5.25)

  
  

• For a plane wave, Eq. 5.11 reveals that $w_E = w_B$, and gives several alternative forms of the energy density
  

  $$w = \frac{B^2}{\mu_o} = \epsilon_o E^2 = \sqrt{\frac{\epsilon_o}{\mu_o}} EB$$ (5.26)

  
  
  which averages to
  

  $$\bar{w} = \frac{B_o^2}{2\mu_o} = \frac{1}{2} \epsilon_o E_o^2 = \frac{1}{2} \sqrt{\frac{\epsilon_o}{\mu_o}} E_oB_o$$ (5.27)

  
  

5.4.2 Intensity and the Poynting Vector

• As with acoustic plane waves (see Eq. 4.35), the intensity is defined as the power carried by the wave per unit area transverse to $\vec{k}$ and is related to the energy density via
  

  $$I = cw$$ (5.28)

  
  
  and
  

  $$\bar{I} = c\bar{w}$$ (5.29)

  
  

• Noting that we expect the power carried by the wave to travel along $\vec{k}$, Eq. 5.26 and 5.28 suggest that we can define an intensity vector
  

  $$\vec{S} = \frac{1}{\mu_o} \vec{E}\times\vec{B}$$ (5.30)

  
  
  with magnitude $S=I$, called the poynting vector.

• The power delivered to an area $\vec{A}$, which is not necessarily transverse to $\vec{k}$, is given by
  

  $$\frac{dU}{dt} = \int^A \vec{S} \cdot \vec{da}$$ (5.31)

  
  

5.4.3 Radiation Pressure

• You learned in your modern physics course that the momentum carried by a photon traveling in vacuum is^5.4
  

  $$p = \frac{E}{c}$$ (5.32)

  
  
  where $E$ is its energy.

• Consider an electromagnetic wave incident normally on a perfect absorber of area $A$. Eq. 5.32 enables us to write Eq. 5.31 as
  

  $$\frac{dp}{dt} = F = \frac{1}{c} \int^A S da$$ (5.33)

  
  

• Differentiating with respect to area and recognizing $\frac{dF}{da}$ as the pressure and the magnitude of $\vec{S}$ as $I$, we have
  

  $$P = \frac{I}{c}$$ (5.34)

  
  

• You will consider the radiation pressure exerted by the sun on objects in our solar system for homework.

Copyright © 2003-2009, Lewis A. Riley

Updated Wed Jan 18 09:51:28 2006