Electrostatics

JEE Main level, numerical answer, medium

294 of 311

Question 294 of 311

The electric field intensity is produced by the radiation coming from a 100W100W bulb at a distance of 3m3m is EE. The electric field intensity produced by the radiation coming from 60W60W at the same distance is x5E\sqrt{\dfrac{x}{5}}E. Where the value of x=x=_________ .

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Answer: 3

cϵ0E2=1004π×32c{\epsilon }_{0}{E}^{2}=\dfrac{100}{4\pi \times {3}^{2}}
⇒cϵ0(x5E)2=604π×32\Rightarrow c{\epsilon }_{0}{\left(\sqrt{\dfrac{x}{5}}E\right)}^{2}=\dfrac{60}{4\pi \times {3}^{2}}
⇒x5=35\Rightarrow \dfrac{x}{5}=\dfrac{3}{5}
⇒x=3\Rightarrow x=3

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