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Suppose I have a simple DC circuit consisting of a battery of voltage $V$ (no internal resistance assumed), and a resistor (resistance $R$) connected in series by conducting wires. Because the wires are held at a potential difference of V and because they have a non-zero capacitance $C_{wires}$, the wire connected to the positive terminal must be positively charged with a charge $Q$ (such that $C_{wires}=\frac{Q}{V}$) and the negative wire must be negatively charged with a charge $-Q$.

But now suppose that as the circuit operates and current flows through the resistor, we ground the negative wire by connecting it to the earth (which we can approximate as an infinitely large conductor with an infinitely large capacitance). This connection means that the negative wire and the earth will try to attain a charge configuration such that there is no potential difference between the earth and the negative wire. But surely the fact that the earth has a virtually infinite capacitance means that a virtually infinite amount of charge (and hence current) will have to be drawn from the battery to charge up the earth to the same potential as the wire? Obviously this doesn't happen otherwise the neutral wire in a household plug would not be able to be grounded. But what does actually happen? How does the grounding of the negatively charged wire affect the charge distribution of the circuit/earth system and why doesn't the earths huge capacitance imply that a huge amount of charge must be required to charge it up to the potential of the negative wire?

There is a similar question as well as an answer here Surface charges on grounded conductors in circuits and energy transfer however despite reading through it, I still don't understand how the grounding of the circuit doesn't change the capacitance of the circuit.

Any help on this issue would be most appreciated!

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How does the grounding of the negatively charged wire affect the charge distribution of the circuit/earth system and why doesn't the earths huge capacitance imply that a huge amount of charge must be required to charge it up to the potential of the negative wire?

You have this backwards. The tiny amount of charge on the wire goes to the earth to bring the wire to the same voltage as the earth. The earth's potential is nearly unaffected.

One thing that you may not recognize is that there are two types of capacitance that are important in this problem. There is the capacitance between the two wires, this is called mutual capacitance and I believe that it is what you were describing. However, there is also the so-called self capacitance. This is the capacitance of each wire with respect to ground. So, before the circuit is grounded you can think of it like this:

Ungrounded circuit

The mutual capacitance, C1, is small, but the self-capacitances, C2 and C3, are even smaller. Note that even though you have not deliberately grounded the circuit, the self capacitance provides an unavoidable capacitative connection to ground.

When you deliberately connect the ground to the wire then the circuit changes to look like this:

Grounded circuit

Then all that does is to short out the very tiny amount of charge on C2. Everything else stays the same and only the minuscule amount of charge on C2 needs to change.

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  • $\begingroup$ Thanks! This is a brilliant answer. Just 2 quick questions to clarify everything. First, does this mean that the negative wire becomes totally neutral upon grounding it? My intuition tells me that it does not become totally neutral because there is still a very small amount of negative charge on the negative plate of C1. Second, Is there a reason the capacitance between the earth and the wires (C2 and C3) is so low despite the vast size of the earth? I understand that C1 is basically negligible for a simple low voltage circuit but am not sure why C2 and C3 would be tiny as well $\endgroup$ Apr 20 at 16:16
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    $\begingroup$ @SalahTheGoat Your intuition is correct. We are using both C1 and C2 to describe the wire so the wire isn't entirely C1 and the wire isn't entirely C2. C2 represents the portion of the charge that is drained off by connecting it to ground. To get values for capacitance you can use this website: emisoftware.com/calculator/wire-over-ground-plane-capacitance for a pair of wires 5 cm long, 1 mm diameter, and 2 mm separation the mutual capacitance is 1E-12 F. For a wire 5 cm long, 1 mm diameter, 1 m away from ground the self capacitance is 3E-13 F (wire over ground plate) $\endgroup$
    – Dale
    Apr 20 at 16:57
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    $\begingroup$ Okay thanks so much! All clear now. Answers like yours are what make this site so great $\endgroup$ Apr 21 at 5:19

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