
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
219.6k+ 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
Electricity and Magnetism Explained: Key Concepts & Applications

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

States of Matter Chapter For JEE Main Chemistry

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Understanding Uniform Acceleration in Physics

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Atomic Structure for Beginners

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

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Centrifugal Force in Physics

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Degree of Dissociation: Meaning, Formula, Calculation & Uses

