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Question: Show that the average energy density of the \(E\) field equals the average energy density of the \(B...

Show that the average energy density of the EE field equals the average energy density of the BB field.

Explanation

Solution

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 gg 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 UU 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, UB=12μ0B2{U_B} = \dfrac{1}{{2{\mu _0}}}{B^2}
For average energy density, we have
UE=12ε0E02{{\text{U}}_{\text{E}}} = \dfrac{1}{2}{\varepsilon _0}{{\text{E}}_0}^2
E0B0=C\Rightarrow \dfrac{{{E_0}}}{{{B_0}}} = C
On substitution we have
UE=14ε0C2  B02{{\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
C=1μ0E0{\text{C}} = \dfrac{1}{{\sqrt {{\mu _0}{{\text{E}}_0}} }}
Upon substitution we get,
UE=14ε0  B021μ0ε0{{\text{U}}_{\text{E}}} = \dfrac{1}{4}{\varepsilon _0}\;{\text{B}}_0^2 \cdot \dfrac{1}{{{\mu _0}{\varepsilon _0}}}
UE=14BO2μ0 UE=(Bo22μ0)=UB\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.