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
High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET
🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
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)
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