Electric field of electromagnetic wave is given by E=50 sint−xcN/C. Then intensity of waves is

I=energy density × C=12 ε0 Em2 × C =12 × 8.85 × 10−12502 × 3 ×108 w/m2=3.3 W/m2

Electric field of electromagnetic wave is given by E=50 sint−xcN/C. Then intensity of waves is

I=energy density × C=12 ε0 Em2 × C =12 × 8.85 × 10−12502 × 3 ×108 w/m2=3.3 W/m2

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Electric field of electromagnetic wave is given by $E = 50 \sin (t - \frac{x}{c}) N / C .$ Then intensity of waves is

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$6 .5 w / m^{2}$

$1 .2 w / m^{2}$

$3 .3 w / m^{2}$

$2 .2 w / m^{2}$

answer is C.

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$I = (energy density) \times C = \frac{1}{2} ε_{0} E_{m}^{2} \times C = \frac{1}{2} \times 8 .85 \times 10^{- 12} {(50)}^{2} \times 3 \times 10^{8} w / m^{2} = 3 .3 W / m^{2}$

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Electric field of electromagnetic wave is given by E=50 sint−xcN/C. Then intensity of waves is

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Electric field of electromagnetic wave is given by $E = 50 \sin (t - \frac{x}{c}) N / C .$ Then intensity of waves is