Electromagnetic Waves

JEE Main level, single correct, medium

127 of 165

Question 127 of 165

An electromagnetic wave of frequency 1×10141 \times 10^{14} hertz is propagating along z\mathrm{z}-axis. The amplitude of electric field is 4 V/m4 \mathrm{~V} / \mathrm{m}. If ε0=8.8×10−12C2/N−\varepsilon_0=8.8 \times 10^{-12} \mathrm{C}^2 / \mathrm{N}- m2\mathrm{m}^2, then average energy density of electric field will be:

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Answer: (C) 35.2×10−12 J/m335.2 \times 10^{-12} \mathrm{~J} / \mathrm{m}^3

Given: Amplitude of electric field,
E0=4v/mE_0=4 \mathrm{v} / \mathrm{m}
Absolute permitivity,
ε0=8.8×10−12c2/N−m2\varepsilon_0=8.8 \times 10^{-12} \mathrm{c}^2 / \mathrm{N}-\mathrm{m}^2
Average energy density uE=u_E= ?
Applying formula,
Average energy density uE=14ε0E2u_E=\frac{1}{4} \varepsilon_0 E^2
⇒uE=14×8.8×10−12×(4)2=35.2×10−12 J/m3\begin{aligned} &\Rightarrow \quad u_E=\frac{1}{4} \times 8.8 \times 10^{-12} \times(4)^2 \\ &=35.2 \times 10^{-12} \mathrm{~J} / \mathrm{m}^3 \end{aligned}

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