
The rms value of the electric field of a plane electromagnetic wave is 314 V / m. The average energy density of electric field and the average energy density are:
(A) $4.3{ \times ^{ - 7}}J{m^{ - 3}}:8.6 \times {10^{ - 7}}J{m^{ - 3}}$
(B) $4.1{ \times ^{ - 7}}J{m^{ - 3}}:8.6 \times {10^{ - 7}}J{m^{ - 3}}$
(C) $2.15{ \times ^{ - 7}}J{m^{ - 3}}:8.6 \times {10^{ - 7}}J{m^{ - 3}}$
(D) $8.6{ \times ^{ - 7}}J{m^{ - 3}}:8.6 \times {10^{ - 7}}J{m^{ - 3}}$
Answer
138.9k+ views
Hint We should know that the electric field is defined as the electric force which is represented per unit charge. Based on the definition we need to define the terms average energy density of the electric field and the average energy density and develop the formula. Once we express the formula we have to put the values from the question to evaluate the expressions.
Complete step by step answer
We should know that the value to find the average energy density of electric field is given by:
$\dfrac{1}{2}{\varepsilon _0}{\varepsilon _{rms}}^2 = 0$
Now we have to put the values from the question into the above expression, to get:
$\dfrac{1}{2} \times 8.85 \times {(314)^2} \times {10^{ - 12}}$
$= 4.36 \times {10^{ - 7}}$
It is also known to us that the average energy density is given up:
${\varepsilon _0}{\varepsilon _{rms}}^2$
Now we have to put the values from the question into the above expression, to get:
$8.72 \times {10^{ - 7}}$.
So, the average energy density of electric field and the average energy density are $4.3{ \times ^{ - 7}}J{m^{ - 3}}:8.6 \times {10^{ - 7}}J{m^{ - 3}}$ .
Hence the correct answer is option A.
Note We should know that the total energy is defined as the sum of all the densities that are associated with the electric and magnetic fields. Whenever we have to take into consideration the instantaneous energy density that is over one or more cycles concerned with an electromagnetic wave, we have to take again so as to get a factor of 1 /2 from the time average from the equation.
The basic definition of the energy density is defined as the amount of energy that will be stored in a given system or region of the space per unit volume.
Complete step by step answer
We should know that the value to find the average energy density of electric field is given by:
$\dfrac{1}{2}{\varepsilon _0}{\varepsilon _{rms}}^2 = 0$
Now we have to put the values from the question into the above expression, to get:
$\dfrac{1}{2} \times 8.85 \times {(314)^2} \times {10^{ - 12}}$
$= 4.36 \times {10^{ - 7}}$
It is also known to us that the average energy density is given up:
${\varepsilon _0}{\varepsilon _{rms}}^2$
Now we have to put the values from the question into the above expression, to get:
$8.72 \times {10^{ - 7}}$.
So, the average energy density of electric field and the average energy density are $4.3{ \times ^{ - 7}}J{m^{ - 3}}:8.6 \times {10^{ - 7}}J{m^{ - 3}}$ .
Hence the correct answer is option A.
Note We should know that the total energy is defined as the sum of all the densities that are associated with the electric and magnetic fields. Whenever we have to take into consideration the instantaneous energy density that is over one or more cycles concerned with an electromagnetic wave, we have to take again so as to get a factor of 1 /2 from the time average from the equation.
The basic definition of the energy density is defined as the amount of energy that will be stored in a given system or region of the space per unit volume.
Recently Updated Pages
Young's Double Slit Experiment Step by Step Derivation

How to find Oxidation Number - Important Concepts for JEE

How Electromagnetic Waves are Formed - Important Concepts for JEE

Electrical Resistance - Important Concepts and Tips for JEE

Average Atomic Mass - Important Concepts and Tips for JEE

Chemical Equation - Important Concepts and Tips for JEE

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Physics Average Value and RMS Value JEE Main 2025

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

Charging and Discharging of Capacitor
