JEE Main 2014 · 11 Apr · Shift 1 · Physics Q22

← JEE Main 2014 · 11 Apr · Shift 1

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An electromagnetic wave of frequency $1\times10^{14}$ hertz is propagating along z - axis. The amplitude of electric field is 4 V/m. If $\epsilon_o=8.8\times10^{-12}$ C $^2$ /N-m $^2$ , then average energy density of electric field will be :

35.2\times10^{-10}

J/m

^3

35.2\times10^{-11}

J/m

^3

35.2\times10^{-12}

J/m

^3

35.2\times10^{-13}

J/m

^3

Answer and worked solution

Official answer: C

$\langle u_E\rangle=\tfrac14\epsilon_0E_0^2=\tfrac14\times8.8\times10^{-12}\times16=35.2\times10^{-12}\text{ J/m}^3.$

Topics:Energy and momentum of EM waves

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