Electronic – The effect of grounding a simple circuit

capacitorchargegroundgroundingvoltage

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 https://physics.stackexchange.com/questions/397545/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!

Best Answer

The Earth's huge capacitance means that it can exchange a lot of charge without changing potential. Your circuit, however, has a comparatively tiny capacitance. If some charge is added or taken, it changes potential a lot (compared to the change in the potential of the Earth). So it is not the Earth changing its potential to meet the circuit, it is the reverse -- the circuit changes its potential to match that of the Earth.

More formally, we should really consider the capacitance of the Earth-circuit system and ask how much the voltage between the two changes if some charge is moved from one to the other. However, the change in the electric field between them, which determines the change in voltage, is all due to the change in the electric field due to the charge transfer to or from the circuit rather than from the Earth.