Verified Solution • 2025–26 Session moderate

QJEEM21Ph903d: JEE Main 2021 Physics - Class 12 JEEM21Ph903d Solution: The electric field in an electromagnetic wave is given by E =…

The electric field in an electromagnetic wave is given by E = (50 NC −- 1) sin ω\omega (t −- x/c)

The energy contained in a cylinder of volume V is 5.5 ×\times 10 −- 12 J. The value of V is _____________ cm3. (given ∈\in 0 = 8.8 ×\times 10 −- 12C2N −- 1m −- 2)

Step-by-Step Verified Solution

E=50sin⁡(ωt−ωc. x)E = 50\sin \left( {\omega t - {\omega \over c}.\,x} \right)

Energy density = 12∈0E02{1 \over 2}{ \in _0}E_0^2

Energy of volume V=12∈0E02. V=5.5×10−12V = {1 \over 2}{ \in _0}E_0^2.\,V = 5.5 \times {10^{ - 12}}

128.8×10−12×2500V=5.5×10−12{1 \over 2}8.8 \times {10^{ - 12}} \times 2500V = 5.5 \times {10^{ - 12}}

V=5.5×22500×8.8=.0005m3V = {{5.5 \times 2} \over {2500 \times 8.8}} = .0005{m^3}

= .0005 ×\times 106 (c.m)3

= 500 (c.m)3

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