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Electrostatics

JEE Main level, numerical answer, medium

107 of 311

Question 107 of 311

The peak electric field produced by the radiation coming from the $8W$ bulb at a distance of $10m$ is $\dfrac{x}{10}\sqrt{\dfrac{{\mu }_{0}c}{\pi }}V{m}^{-1}$ . The efficiency of the bulb is $10%$ and it is a point source. The value of $x$ is,

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Answer: 2

I=\dfrac{1}{2}c\epsilon E020

\dfrac{8}{4\pi \times {10}^{2}}\times \dfrac{1}{2}=\dfrac{1}{4}\times c\times \dfrac{1}{{\mu }_{0}{c}^{2}}\times {E}_{0}^{2}

{E}_{0}=\dfrac{2}{10}\times \sqrt{\dfrac{{\mu }_{0}c}{\pi }}\Rightarrow x=2

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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JEE Main Physics practice question on Electrostatics (Electric Field and Electric Field Lines), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.

The peak electric field produced by the radiation coming from the 8W8W8W bulb at a distance of 10m10m10m is x10μ0cπVm−1\dfrac{x}{10}\sqrt{\dfrac{{\mu }_{0}c}{\pi }}V{m}^{-1}10x​πμ0​c​​Vm−1. The efficiency of the bulb is 1010%10 and it is a point source. The value of xxx is,

JEE Main question paper PDF

More Physics practice

The peak electric field produced by the radiation coming from the $8W$ bulb at a distance of $10m$ is $\dfrac{x}{10}\sqrt{\dfrac{{\mu }_{0}c}{\pi }}V{m}^{-1}$ . The efficiency of the bulb is $10%$ and it is a point source. The value of $x$ is,