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Show that the average energy density of the $E$ field equals the average energy density of the $B$ field.

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Last updated date: 17th Jul 2024
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Answer
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Hint: The ratio of the amount of electric charge deposited on a conductor to the difference in electric potential is known as capacitance. Self capacitance and reciprocal capacitance are two closely related concepts of capacitance. Self capacitance is a property of any material that can be electrically charged.

Complete step by step answer:
The sum of energy contained in a given structure or area of space per unit volume is referred to as energy density in physics. It may also refer to energy per unit mass, but real energy is a more precise expression (or gravimetric energy density). The cumulative sum of energy in a device per unit volume is known as energy density.

The amount of $g$ of sugar in food, for example. Low energy dense foods have less calories per gram of food, allowing you to consume more of them so there are less calories. The letter $U$ is used to represent it. Magnetic and electric fields have the ability to accumulate energy as well.

As it comes to electromagnetic waves, both the magnetic and electric fields play a role in determining energy density. As a result, the expression for energy density is the sum of the electric and magnetic field's energy density. For average magnetic density we have
Since, ${U_B} = \dfrac{1}{{2{\mu _0}}}{B^2}$
For average energy density, we have
${{\text{U}}_{\text{E}}} = \dfrac{1}{2}{\varepsilon _0}{{\text{E}}_0}^2$
$\Rightarrow \dfrac{{{E_0}}}{{{B_0}}} = C$
On substitution we have
${{\text{U}}_{\text{E}}} = \dfrac{1}{4}{\varepsilon _0} \cdot {{\text{C}}^2}\;{{\text{B}}_0}^2$
We know that the speed of Electromagnetic waves is
${\text{C}} = \dfrac{1}{{\sqrt {{\mu _0}{{\text{E}}_0}} }}$
Upon substitution we get,
${{\text{U}}_{\text{E}}} = \dfrac{1}{4}{\varepsilon _0}\;{\text{B}}_0^2 \cdot \dfrac{1}{{{\mu _0}{\varepsilon _0}}}$
$\Rightarrow {{\text{U}}_{\text{E}}} = \dfrac{1}{4}\dfrac{{{\text{B}}_{\text{O}}^2}}{{{\mu _0}}} \\
\therefore {{\text{U}}_{\text{E}}}= \left( {\dfrac{{{{\text{B}}_o}^2}}{{2{\mu _0}}}} \right) = {U_{\text{B}}}$

Hence, the average energy density of the E field equals the average energy density of the B field.

Note: Electromagnetic waves, or EM waves, are waves that are produced when an electric field and a magnetic field vibrate together. EM waves, in other words, are made up of oscillating magnetic and electric fields. Electromagnetic radiation is a term used in physics to describe the waves of the electromagnetic field that propagate through space and carry electromagnetic radiant energy. Radio waves, microwaves, infrared, sun, ultraviolet, X-rays, and gamma rays are also examples of electromagnetic radiation. The electromagnetic spectrum includes both of these waves.