This is a dumb question. I recently found out that earth has an electric field around 150V/m near it's surface. So,I was wondering if we held the two terminals of the bulb few centimetres apart, verticaly above the ground ,shouldn't the bulb shine? I know this doesn't happen but, I am unable to comprehend this phenomenon. Why can't the earth's electric field generate a potential difference between bulb's terminals?
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$\begingroup$ Did you move it? $\endgroup$– Charles Tucker 3Commented Aug 5, 2022 at 7:27
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$\begingroup$ is this what you are talking about? en.wikipedia.org/wiki/Atmospheric_electricity#Description . Agnius answers it. $\endgroup$– anna vCommented Aug 5, 2022 at 7:58
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$\begingroup$ You may find it helpful to read the article: "9-1 The electric potential gradient of the atmosphere" in the Feynman lecture's on Physics Vol 2. The bulb and the wire become equipotential $\phi=constant$. $\endgroup$– HEMMICommented Aug 5, 2022 at 8:22
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$\begingroup$ Here is a link to the Feynman's chapter Electricity in the Atmosphere mentioned by @HEMMI. $\endgroup$– FarcherCommented Aug 5, 2022 at 9:09
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$\begingroup$ @CharlesTucker3 I actually doubted that. So you are saying if we move the whole set in a sufficient speed, it would make a flow of current and therefore the the bulb will glow, right? $\endgroup$– donthababakkaCommented Aug 5, 2022 at 13:04
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2 Answers
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Local atmospheric current densities are very weak, about $10^{-12}~\text {A}/\text{m}^2$, so per electric power law :
$$ P = VI ,$$
few centimeters apart between two terminals of bulb, would generate only about $\approx 10^{-12}~\text{W}$ of electric power, which clearly is not enough to light-up bulb with nominal $100~\text{W}$ power. Thus, unless bulb somewhat near lightning zone,- it will never happen.
answered Aug 5, 2022 at 7:29
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A light bulb pops into existence with one of its terminals in the air 1m above the ground and the other terminal in the air 1.05m above the ground, so that there is a potential difference of ~5V across the terminals. What happens right away? A very small amount of charge passes through the light bulb from one terminal to the other (and maybe lights it slightly for a microsecond). Now the voltage across the terminals is 0V. The air immediately next to one terminal is at the same potential as the air immediately next to the other terminal.
What happens after that? Pretty much nothing, because air is basically an insulator. Nothing is going to come along and push more current through the bulb.
The processes that ionize the atmosphere, creating that natural electric field, produce a very, very small current. It's only because air is a very poor conductor that that tiny current is able to build up a significant voltage. As soon as you introduce something that's a better conductor than air, that tiny current develops a tiny voltage, and an even tinier power.
