What does earthing actually do in electrostatics? As far as i know, it simply sets the potential of the object that is earthed to zero.
To explain my question further with an example, consider this question i had to solve recently:
In the diagram shown, we have three large, identical parallel conducting plates A, B and C placed such that the switches $S_1$ and $S_2$ are open initially, and they can be used to earth the plates A and C just by closing them. A charge $+Q$ is given to the plate B. It is observed that a charge of amount:
(A) $Q$ will pass through $S_1$, when $S_1$ is closed and $S_2$ is open
(B) $Q$ will pass through $S_2$, when $S_2$ is closed and $S_1$ is open
(C) $Q/3$ will pass through $S_1$, $2Q/3$ will pass through $S_2$, when $S_1$ and $S_2$ are both closed together
(D) $4Q/3$ will pass through $S_1$, $-Q/3$ will pass through $S_2$, when $S_1$ and $S_2$ are both closed together
The correct options are given to be (A), (B) and (C). Applying my definition, i found that the option (C) is correct by making the potentials of the plates A and C due to the other two equal to zero. But i am struggling to do so when only one of the plates is grounded. Whatever charge may flow from the earthed plate, it should not affect the potential of the plate itself (does it?). It is influenced only by the other plates and the charges on them.
This begs the question, what actually happens on earthing something in electrostatics?
Also, how can we solve it without using the capacitor treatment? I am trying to understand the underlying physics of it!
Thanks.
Best Answer
The credit to this answer goes to @knzhou, who provided it to me in the chat.
What earthing does is simply equalizes the potential of the Earth (as in the planet, or the ideal earth, whichever is relevant). Now, we see that to calculate the potential difference between the plate (or whatever it is to which the earth is connected) and any point on or in the earth, we integrate the field as: $$ V_{\text{plate}}-V_{\text{earth}}=-\int_{\text{earth}}^{\text{plate}}\vec{E\,}\cdot\mathrm d\vec{r\,}$$
There are many points which lie on the earth, and hence for the potential difference to always be zero, the electric field (outside of the region between two plates) must always be zero. For the example discussed, this is only possible if the plate A gets a negative charge $-Q$, which it takes from the earth itself.