Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

An electromagnetic wave of frequency 1×1014 hertz is propagating along z-axis. The amplitude of electric field is 4 V/m. If ε0=8.8×10-12C2/N-m2, then average energy density of electric field will be:

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

a

35.2×10-10 J/m3

b

35.2×10-11 J/m3

c

35.2×10-12 J/m3

d

35.2×10-13 J/m3

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

Given: Amplitude of electric field,  E0=4v/m  Absolute permitivity,  ε0=8.8×10-12c2/N-m2  Average energy density uF=?  Applying formula,   Average energy density uE=14ε0E2 ⇒  uE=14×8.8×10-12×(4)2 =35.2×10-12 J/m3
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring