Question

SHOW THAT THE AVERAGE ENERGY DENSITY OF THE "E" FIELD EQUALS THE AVERAGE ENERGY DENSITY OF THE "B" FIELD.
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Answer 1

Answer:

Solution :

Average energy density of electric field is given

by,

uE=12∈0E2rms=12∈0(E02–√)2=14∈0E20....(i)uE=12∈0Erms2=12∈0(E02)2=14∈0E02....(i)

Average energy density of magnetic field is given by,

uB=12μ0B2rms=12μ0(B02–√)2=14B20μ0.....(ii)uB=12μ0Brms2=12μ0(B02)2=14B02μ0.....(ii)

We know E0=cB0orB0=E0/cE0=cB0orB0=E0/c

Also c=1μ0∈0−−−−−√c=1μ0∈0

Putting values in (ii) , we get

uB=14μ0E20c2=14μ0(E20)×μ0∈0=14∈0E20=uE

Explanation:

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Answer 2

Answer:

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We have to show that average energy density of the electric field = average energy density of the magnetic field

we know, average energy density in electric field , U_E=\frac{1}{2}\epsilon_0E^2......(1)

where, E is electric field intensity and \epsilon_0 is permittivity of medium.

and average energy density in magnetic field , U_M=\frac{1}{2}\frac{B^2}{\mu_0}........(2)

where B is magnetic field intensity and \mu_0 is permeability of medium.

also we know, E=cB and c^2=\frac{1}{\mu_0\epsilon_0}, putting these in equations (1) and (2),

we get, U_E=U_M

hence, average energy density in electric field = average energy density in magnetic field .