# Proof of charge existence on a grounded conductor

A question regarding the existence of charge on grounded conductors is confusing me.

Could there be charge on a grounded conductor? How does this not contradict Gauss's Law?

Since every conductor has some capacitance attributed to it, doesn't the existence of charge on a grounded conductor contradicts the equation: $Q = C*V$ ?

-

"ground" is an ambiguity... It's where you define V to be zero... But the physical quantity is Potential difference. – Chris Gerig Aug 9 '12 at 22:18
– Qmechanic♦ Aug 9 '12 at 22:51
The answers doesn't provide a proof, nor does it answer the capacitor's equation – alqubaisi Aug 9 '12 at 23:58
What does Gauss's law have anything to do with zero potential, i.e., grounded? – C.R. Sep 9 '12 at 15:16

## 2 Answers

It is almost guaranteed that many macroscopic earth-grounded conductors are not exactly charge-neutral. But they are probably all extremely close to charge-neutral.

The reason is, "ground" has no divine status, it is just a conductor. Usually it's earth-ground, and earth is very close (I imagine) to charge-neutral. But imagine life on a planet that was quite negatively charged (due to some atmospheric process throwing protons but not electrons into space). Then if you connected a conductor to "earth-ground" (i.e. the planet) it would certainly take some of the excess electrons.

You are wondering how the equation Q=CV applies to a capacitor where both sides are charged and both sides are grounded. Well, for the equation Q = CV, Q is not just any old charge, it refers specifically to the so-called "charge on the capacitor", the amount of charge that has been added to one plate but subtracted from the other plate. If the same charge is added to both plates, that charge is not "the charge on the capacitor", it's actually the charge on a different capacitor, the capacitor where one "plate" is both of those conductors, and the other "plate" is wherever the counter-charge is (i.e. the other end of those electric field lines), which would be outer space. So there is no contradiction: The two grounded plates indeed have no voltage difference between them, Q=V=0 in the equation.

For Gauss's law, I'm not sure what you think the contradiction is. There are electric field lines going from space to the planet's surface and to grounded conductors. On the other hand, a grounded conductor in a grounded metal building really would be charge-neutral, because the electric field lines from outer space would terminate on the outside of the building and would not get inside.

-

The earth can be considered like a large spheric conductor of capacity C1. When you ground something whose capacity is C2, the electric charges move from one conductor to the other: The equations are:

Q1+Q2=Q'1+Q'2

V'1=V'2=V

Q'1=C1*V Q'2=C2*V

The Earth electric capacity is so huge that unless you do microelectronics, you can consider all the charges have flown to it.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9536361694335938, "perplexity_flag": "middle"}