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The average energy density associated with electric field $E = 20 \sin (100 t) N / C$ is

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$6 .98 \times 10^{- 4} J / m^{3}$

$8 .21 \times 10^{- 21} J / m^{3}$

$8 .85 \times 10^{- 10} J / m^{3}$

$16 .9 \times 10^{- 6} J / m^{3}$

answer is C.

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Detailed Solution

$\begin{array}{l} < u >= \frac{1}{2} ε_{0} E_{rms}^{2} \\ < u >= \frac{1}{2} ε_{0} {(\frac{20}{\sqrt{2}})}^{2} \\ = \frac{8 .85 \times 10^{- 12}}{2} \times 200 \\ = 8 .85 \times 10^{- 10} J / m^{3} \end{array}$

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