Capacitors

JEE Main level, single correct, medium

130 of 180

Question 130 of 180

The electric field in an electromagnetic wave is given as
E⃗=20sin⁡ω(t−xc)j⃗NC−1\vec{E}=20\sin \omega \left(t-\dfrac{x}{c}\right)\vec{j}N{C}^{-1}, where ω\omega and cc are angular frequency and velocity of electromagnetic wave respectively. The energy contained in a volume of 5×10−4m35\times {10}^{-4}{m}^{3} will be
(Given ϵ0=8.85×10−12C2N−1m−2{\epsilon }_{0}=8.85\times {10}^{-12}{C}^{2}{N}^{-1}{m}^{-2})

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Answer: (D) 8.85×10−13J8.85\times {10}^{-13}J

Given that, E⃗=20sin⁡ω(t−xc)j⃗NC−1\vec{E}=20\sin \omega \left(t-\dfrac{x}{c}\right)\vec{j}N{C}^{-1}
The average energy density of an electromagnetic wave is 12ϵ0E20\dfrac{1}{2}{\epsilon }_{0}{{E}^{2}}_{0}.
The total energy stored is
Etotal=12ϵ0E02×volume{E}_{\text{total}}=\dfrac{1}{2}{\epsilon }_{0}{E}_{0}^{2}\times volume
⇒Etotal=12×8.85×10−12×(20)2×5×10−4=8.85×10−13J\Rightarrow {E}_{total}=\dfrac{1}{2}\times 8.85\times {10}^{-12}\times (20{)}^{2}\times 5\times {10}^{-4} =8.85\times {10}^{-13}J

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