An electromagnetic wave of frequency 1 × 1014 Hz is propagating along z-axis. The amplitude of the electric field is 4 V / m. If ε0=8.8 × 10-12 C 2 / N - m 2, then the average energy density of electric field will be :

Question

An electromagnetic wave of frequency 1 × 1014 Hz is propagating along z-axis. The amplitude of the electric field is 4 V / m. If ε0=8.8 × 10-12 C 2 / N - m 2, then the average energy density of electric field will be : (A) 35.2 × 10-12 J / m 3 (B) 35.2 × 10-10 J / m 3 (C) 35.2 × 10-11 J / m 3 (D) 35.2 × 10-13 J / m 3

Tardigrade Admin · Accepted Answer

f =1014 Hz E 0=4 V / m E 0=8.8 × 10-12 C 2 / N - m 2 Energy density of electric field =1 / 2 (Total energy density) =(1/2) ⋅ (1/2) ε0 E 2 =(1/2) ⋅ (1/2) 8.8 × 10-12 × 42 =35.2 × 10-12 J / m 3

Q. An electromagnetic wave of frequency $1 \times 1 0^{14} Hz$ is propagating along z-axis. The amplitude of the electric field is $4 V / m$ . If $ε_{0} = 8.8 \times 1 0^{- 12} C^{2} / N - m^{2}$ , then the average energy density of electric field will be :

$35.2 \times 1 0^{- 12} J / m^{3}$

$35.2 \times 1 0^{- 10} J / m^{3}$

$35.2 \times 1 0^{- 11} J / m^{3}$

$35.2 \times 1 0^{- 13} J / m^{3}$

Q. An electromagnetic wave of frequency 1×1014Hz is propagating along z-axis. The amplitude of the electric field is 4V/m. If ε0​=8.8×10−12C2/N−m2, then the average energy density of electric field will be :

35.2×10−12J/m3

35.2×10−10J/m3

35.2×10−11J/m3

35.2×10−13J/m3

f=1014Hz E0​=4V/m E0​=8.8×10−12C2/N−m2 Energy density of electric field =1/2 (Total energy density) =21​⋅21​ε0​E2 =21​⋅21​{8.8×10−12×42} =35.2×10−12J/m3

Q. An electromagnetic wave of frequency $1 \times 1 0^{14} Hz$ is propagating along z-axis. The amplitude of the electric field is $4 V / m$ . If $ε_{0} = 8.8 \times 1 0^{- 12} C^{2} / N - m^{2}$ , then the average energy density of electric field will be :

$35.2 \times 1 0^{- 12} J / m^{3}$

$35.2 \times 1 0^{- 10} J / m^{3}$

$35.2 \times 1 0^{- 11} J / m^{3}$

$35.2 \times 1 0^{- 13} J / m^{3}$