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Electrostatic Potential And Capacitance

Question
CBSEENPH12039490

A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the diode is:

  • 1.73 V/m

  • 2.45 V/m

  • 5.48 V/m

  • 7.75 V/m

Solution

B.

2.45 V/m

Consider the LED as a point source of light.
Let power of the LED is P

Intensity at r from the source
straight I space equals space fraction numerator straight P over denominator 4 πr squared end fraction ... (i)
As we know that,
straight I space equals space 1 half straight epsilon subscript straight o straight E subscript 0 superscript 2 straight c .... (ii)
From eqs. (i) and (ii) we can write
fraction numerator straight P over denominator 4 πr squared end fraction space equals space 1 half space straight epsilon subscript straight o straight E subscript 0 superscript 2 straight c
straight E subscript 0 superscript 2 space equals space fraction numerator 2 straight P over denominator 4 πε subscript 0 straight r squared straight c end fraction space equals space fraction numerator 2 space straight x space 0.1 space straight x space 9 space straight x space 10 to the power of 9 over denominator 1 space straight x space 3 space straight x space 10 to the power of 8 end fraction
straight E subscript 0 superscript 2 space equals space 6
straight E subscript 0 space equals space square root of 6 space equals space 245 space straight V divided by straight m

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