I don't understand how the average energy density and total radiated power of an infinitesimal dipole above a conductive surface is being calculated and why the magnetic field is not being considered in this calculation.

From Balanis:

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whereas the general definition of the average energy density is $$W_{av}=\frac{1}{2}\mathrm{Re}|E\times H^{*}|^2$$

  • \$\begingroup\$ You need to ask a question and, in this particular case, justify it with a reason why you feel it is wrong. \$\endgroup\$
    – Andy aka
    Mar 5, 2023 at 23:01
  • 2
    \$\begingroup\$ infinitesimal means smaller than microscopic ... is that what you meant to say? \$\endgroup\$
    – jsotola
    Mar 5, 2023 at 23:46
  • \$\begingroup\$ ok I though of the title and explained my question in the text body I will edit it \$\endgroup\$
    – Shakiba HS
    Mar 6, 2023 at 1:29

2 Answers 2


The magnetic field is still being considered in the equation. For example, consider the average Poynting vector

\$ W_{av} = \dfrac{1}{2} Re[E\times H^*] \$

For a plane wave the magnetic field can be related to the electric field through the impedance \$ \eta \$ as

\$ H = \dfrac{1}{\eta} \hat{n} \times E \$ where \$ \hat{n} \$ is the direction of propagation of the wave. Plugging this into the equation for \$ W_{av} \$ gives

\$ W_{av} = \dfrac{1}{2}Re[E\times\dfrac{1}{\eta}\hat{n} \times E^*] = \dfrac{1}{2\eta} Re[E\times E^*]\hat{n} = \dfrac{1}{2\eta}\ |E|^2 \hat{n} \$. There are some additional factors to consider for the infinitesimal dipole but I hope you can see where some of the factors in the final equation are coming from and can see that the magnetic field IS being considered in the equation

  • \$\begingroup\$ yes, I noticed that E/H= intrinsic impedance and I was careless about the etha in the denominator of average energy density thank you. \$\endgroup\$
    – Shakiba HS
    Mar 6, 2023 at 22:27

The key thing to understand here is that for an infinitesimal dipole above a perfect conductor, the electromagnetic fields simplify greatly. In particular:

The magnetic field is identically zero in the region above the conductor. This is because magnetic fields cannot penetrate a perfect conductor. The electric field takes on a very simple form - it is radial and proportional to 1/r3.

So in this special case, the average energy density and radiated power can be calculated from just the electric field, without considering the magnetic field (since it is zero). The expression from Balanis is calculating the average energy density from just the electric field term (|E|2) since the magnetic term (H∗) drops out. In general, for arbitrary sources and fields, you do need to consider both E and H - but for this specific case of a dipole above a conductor, the fields are highly simplified and the magnetic portion is not needed.

  • \$\begingroup\$ Warning: Systems have detected this answer may be AI-generated (e.g. ChatGPT) and therefore may be untrustworthy. Exercise additional caution when relying on its contents. It may be deleted in future. \$\endgroup\$
    – SamGibson
    Mar 6, 2023 at 7:19
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    \$\begingroup\$ This answer is incorrect. If the electric field fell off as \$ \dfrac{1}{r^3} \$ then power wouldn't be radiated since Power will fall off as \$ \dfrac{1}{r^6} \$ \$\endgroup\$
    – CMH12
    Mar 6, 2023 at 15:04

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