Principle of superposition for charges/induced charges

[phantomvommand]

The key observation to solve the above problem is that the charge Q can be dragged out into a flat capacitor plate parallel to the 2 existing plates. Apparently, while the charge distribution on the 2 existing plates changes, the total charge induced on each plate remains the same, due to the principle of superposition.

How and why does the principle of superposition work here??

If I had 2 concentric spherical shells instead of 2 parallel plates, and I placed a charge in the region between the concentric shells, can I likewise drag the charge out into a shell and claim that the total induced charges on each of the 2 existing shells remains the same due to the principle of superposition?

If the answer to the above question is yes, when does the principle of superposition fail? Please do give some examples, that are ideally somewhat similar to the above questions, so as to elucidate the exact criterions that cause the principle of superposition to be inapplicable.

Thank you!

 

[etotheipi]

There are a few things going on here!

Firstly, take a capacitor with plates of infinite extent (say, each parallel to the $xy$ plane). By translational symmetry parallel to the plates it follows the charges $q_A$ and $q_B$ induced on the capacitor plates will not depend on the coordinates $(x,y)$ at which you position an arbitrary charge $q$. For this problem, then, it's necessary to assume that if $A \propto l^2$, then $l \gg d$, or in other words that edge effects are negligible.

Now for the superposition part. Suppose you have a set of charges $\{ Q^{(1)}, \dots, Q^{(N)} \}$. If you put only the charge $Q^{(1)}$ between the plates (at the specified $z$ value, and arbitrary $(x,y)$ coordinates), then the corresponding charge induced on the plates will be $q_A^{(1)}$ and $q_B^{(1)}$. But by superposition, you know that the total charges induced on the plates, $q_A$ and $q_{B}$, will just be the sum of those that would be induced if the $Q^{(i)}$ charges were on their own one at a time, i.e.$$q_{A} = q_A^{(1)} + \dots + q_A^{(N)}, \quad \quad q_{B} = q_B^{(1)} + \dots + q_B^{(N)}$$So the solution to the problem is to note that you can start with a charge $Q$, partition it into $N$ smaller charges and spread them out parallel to the plates to different locations, without affecting the total charge on either plates. Further, by Gauss, you can write down$$q_{A} + q_{B} + Q = 0$$To work out the ratio $q_{A}/q_{B}$, now just use the fact that you have effectively two capacitors in series, with the two outermost plates at the same electric potential!

 

[phantomvommand]

There are a few things going on here!

Firstly, take a capacitor with plates of infinite extent (say, each parallel to the $xy$ plane). By translational symmetry parallel to the plates it follows the charges $q_A$ and $q_B$ induced on the capacitor plates will not depend on the coordinates $(x,y)$ at which you position an arbitrary charge $q$. For this problem, then, it's necessary to assume that if $A \propto l^2$, then $l \gg d$, or in other words that edge effects are negligible.

Now for the superposition part. Suppose you have a set of charges $\{ Q^{(1)}, \dots, Q^{(N)} \}$. If you put only the charge $Q^{(1)}$ between the plates (at the specified $z$ value, and arbitrary $(x,y)$ coordinates), then the corresponding charge induced on the plates will be $q_A^{(1)}$ and $q_B^{(1)}$. But by superposition, you know that the total charges induced on the plates, $q_A$ and $q_{B}$, will just be the sum of those that would be induced if the $Q^{(i)}$ charges were on their own one at a time, i.e.$$q_{A} = q_A^{(1)} + \dots + q_A^{(N)}, \quad \quad q_{B} = q_B^{(1)} + \dots + q_B^{(N)}$$So the solution to the problem is to note that you can start with a charge $Q$, partition it into $N$ smaller charges and spread them out parallel to the plates to different locations, without affecting the total charge on either plates. Further, by Gauss, you can write down$$q_{A} + q_{B} = Q$$To work out the ratio $q_{A}/q_{B}$, now just use the fact that you have effectively two capacitors in series, with the two outermost plates at the same electric potential!

This has been very helpful, thank you very much!