SATHEE: Optics Question 177

Optics Question 177

Question: In an electromagnetic wave, the amplitude of electric field is 1 V/m. the frequency of wave is $5 \times 10^{14} H z$ . The wave is propagating along z-axis. The average energy density of electric field, in Joule/m3, will be

Options:

A) $1.1 \times 10^{- 11}$

B) $2.2 \times 10^{- 12}$

C) $3.3 \times 10^{- 13}$

D) $4.4 \times 10^{- 14}$

Show Answer

Answer:

Correct Answer: B

Solution:

Average energy density of electric field is given by $u_{e} = \frac{1}{2} ε_{0} E^{2} = \frac{1}{2} ε_{0} {(\frac{E_{0}}{\sqrt{2}})}^{2} = \frac{1}{4} ε_{0} E_{0}^{2}$

$= \frac{1}{4} \times 8.85 \times 10^{- 12} {(1)}^{2} = 2.2 \times 10^{- 12} J / m^{3} .$

Optics Question 177

Question: In an electromagnetic wave, the amplitude of electric field is 1 V/m. the frequency of wave is $5 \times 10^{14} H z$ . The wave is propagating along z-axis. The average energy density of electric field, in Joule/m3, will be

Options:

Answer:

Solution:

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Optics Question 177

Question: In an electromagnetic wave, the amplitude of electric field is 1 V/m. the frequency of wave is 5×1014Hz . The wave is propagating along z-axis. The average energy density of electric field, in Joule/m3, will be

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: In an electromagnetic wave, the amplitude of electric field is 1 V/m. the frequency of wave is $5 \times 10^{14} H z$ . The wave is propagating along z-axis. The average energy density of electric field, in Joule/m3, will be