Q9 / 30 · NTA Abhyas (NEET) 2020 · Physics · Electromagnetic Waves

An electromagnetic wave of frequency 1× 10 14  Hz is propagating along z -axis. The amplitude of the electric field is 4 V/m. If ε0=8.8×10-12 C2/N-m2, then the average energy density of electric field will be

  • Correct answer
    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

Solution

f⁡ = 1 0 1 4  Hz
E 0 =4 V/m
∈ 0 = 8.8 × 1 0 - 1 2 C 2 N - m 2
Energy density of electric field
= 1 2 (Total energy density)
= 1 2 · 1 2 ∈ 0 E 2
= 1 2 . 1 2 8.8 × 1 0 - 1 2 × 4 2
= 1 2 · 1 2 ×16×8.8× 10 −12   J/m 3
=35.2× 10 −12   J/m 3

Hence, the correct answer is option: 'A'.

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