Physics · 2014
JEE Main · 11 April 2014, Shift 1 · Q22
An electromagnetic wave of frequency 1 × 10^14 hertz is propagating along z - axis. The amplitude of electric field is 4 V / m. If ε_o =8.8 × 10^-12…
An electromagnetic wave of frequency $\displaystyle 1 \times 10^{14}$ hertz is propagating along z - axis. The amplitude of electric field is $\displaystyle 4 \mathrm{~V} / \mathrm{m}$. If $\displaystyle \epsilon_{\mathrm{o}}=8.8 \times 10^{-12} \mathrm{C}^2 / \mathrm{N}-\mathrm{m}^2$, then average energy density of electric field will be :
Official answer
(3)
$\displaystyle 35.2 \times 10^{-12} \mathrm{~J} / \mathrm{m}^3$
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JEE Main 2014 Physics question, with the answer from NTA’s final answer key. Where our answers come from.