Given: Amplitude of electric field, E0=4v/m Absolute permitivity, ε0=8.8×10−12c2/N−m2 Average energy density uE= ? Applying formula, Average energy density uE=41ε0E2 ⇒uE=41×8.8×10−12×(4)2=35.2×10−12 J/m3
JEE Main 2014 — Physics Electromagnetism
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:
Held on 11 Apr 2014 · Verified 6 Jul 2026.
35.2×10−10 J/m3
35.2×10−11 J/m3
35.2×10−12 J/m3
35.2×10−13 J/m3
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