The electric field in an electromagnetic wave is given by E=50NC−1sinωt−xc . The energy contained in a cylinder of volume V is 5.5×10−12J. The value of V is ……cm3 .(Given ε0=8.85×10−12C2N−1m−2)

E=50sinωt−xcU=5.5×10−12JLet volume = V∈0=8.85×10−12c2m−2N−1We know that, 12∈0E02=UvolumeV=2U∈0E02=2×5.5×10-128.85×10-12×2500=497 cm3

The electric field in an electromagnetic wave is given by E=50NC−1sinωt−xc . The energy contained in a cylinder of volume V is 5.5×10−12J. The value of V is ……cm3 .(Given ε0=8.85×10−12C2N−1m−2)

E=50sinωt−xcU=5.5×10−12JLet volume = V∈0=8.85×10−12c2m−2N−1We know that, 12∈0E02=UvolumeV=2U∈0E02=2×5.5×10-128.85×10-12×2500=497 cm3

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The electric field in an electromagnetic wave is given by $E = (50 {NC}^{- 1}) sinω (t - \frac{x}{c})$ . The energy contained in a cylinder of volume $V$ is $5.5 \times 10^{- 12} J$ . The value of V is …… $c m^{3}$ .
(Given $ε_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ )

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Detailed Solution

$E = 50 \sin ω (t - \frac{x}{c})$

$U = 5 .5 \times 10^{- 12} J$

Let volume = V

$\in_{0} = 8.85 \times 10^{- 12} c^{2} m^{- 2} N^{- 1}$

We know that,

$\frac{1}{2} \in_{0} {E_{0}}^{2} = \frac{U}{v o l u m e}$

$V = \frac{2 U}{\in_{0} {E_{0}}^{2}} = \frac{2 \times 5.5 \times 10^{- 12}}{8.85 \times 10^{- 12} \times 2500} = 497 {cm}^{3}$

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The electric field in an electromagnetic wave is given by E=50NC−1sinωt−xc . The energy contained in a cylinder of volume V is 5.5×10−12J. The value of V is ……cm3 .(Given ε0=8.85×10−12C2N−1m−2)

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The electric field in an electromagnetic wave is given by $E = (50 {NC}^{- 1}) sinω (t - \frac{x}{c})$ . The energy contained in a cylinder of volume $V$ is $5.5 \times 10^{- 12} J$ . The value of V is …… $c m^{3}$ .
(Given $ε_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ )