The electric field intensity produced by the radiation coming from a 100 W bulb at a distance of 3 m is E. The electric field intensity produced by the radiation coming from 60W at the same distance is √(x/5)E. Where the value of x = ____________.

Solution

Physics sir · 3 steps

Intensity goes as E², so the field goes as the square root of the bulb's power.

1. Intensity from power. A bulb spreads its power over a sphere, so at distance r the intensity is I = P/(4π r²). The distance is the same for both bulbs, so I ∝ P.

   I = 1/2ε₀ c E₀² ⇒ E₀ ∝ √I ∝ √P

2. Take the ratio. E'/E = √(60/100) = √(3/5), so E' = √(3/5) E.
3. So. Comparing with √(x/5) E gives x = 3.

The common slip: 1.8. Taking E ∝ P instead of E ∝ √P gives E' = 0.6E, so x/5 = 0.36 and x = 1.8, not 3.

The idea to keep. Energy carried by an EM wave depends on the square of the field. If power or intensity changes by some factor, the field changes by the square root of that factor.

Answer from the NTA final answer key. Solution by PrepWiser.