I’ve seen quite a few questions where some charge is given to a conducting plate and we are to find the final charge on each face of the plate and on the faces of some nearby plates. While there have been some great answers (like this one), I don’t feel the questions were entirely useful to my needs. I would like to build a general method to go about solving such systems (using the basic laws and formulas of electrostatics, of course).

Here are some assumptions:

  1. There are no edge effects
  2. The initial charge distribution is transient

My attempt:

Let’s first assume the basic case of two parallel plates a distance $d$ apart with a charge $q$ given to one and $kq$ to the other.

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Let’s assume $k,q>0$ for simplicity.

The resultant field will cause the charges to shift. Here is the first problem I have. In what configuration will the charges settle? I would imagine all the like charges would want to be as far from each other as possible. Does this mean there would be 0 charge on their inner surfaces, leaving net $q$ and $kq$ for the outer ones respectively?

Once we have determined this, we may consider some other cases.

Considering the case where the plates are given $q$ and $kq$ and they are connected to each other by a conducting wire. First of all, the potentials of both plates have been made equal. So some charge will flow through the wire and the final charges on both plates will be $(k+1)\frac q2$.

enter image description here

But again, the question arises: how much charge is on each surface? Per my understanding of electrostatic shielding, and due to symmetry considerations the charges must be evenly split on the inner and outer surfaces of both plates otherwise the residue charges will cause a field in between the supposedly equipotential surfaces.

And, let’s consider the case where the plate with charge $kq$ is grounded. Again, it seems rather arbitrary to me. All we know is the plate which began with $kq$ has potential $V=0$.

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I suppose that the potential at the right face of the grounded plate should be $0$, and the potential due to the plate with $q$ is $qd\over A\epsilon$ so potential due to charges on the grounded plate should be the same. But I’m unsure how this may be achieved.

I believe these cases are sufficient to solve more complex combinations. However, I am a bit wary of systems where the distance between some pair of plates is different than others. But, I feel it will be worked out with the rest of our cases.


1 Answer 1


First of all, before I explain, we should be clear about a few things.

  • The plates have a finite thickness(although for representative purposes, we draw them using thin lines).
  • The plates are conductors and at any point inside(we need a thick plate to visualise and define the inside) the conductor, the net electric field is zero.
  • The field due an uniformly charged infinite plane at a distance $r$ from it is, $$\mathbf{E}=\frac{\sigma}{2\epsilon_0}\hat{\mathbf{p}}$$ where $\sigma$ is the surface charge density of the plate and $\hat{\mathbf{p}}$ is the normal vector of the infinitely charged plane.(Note that the electric field is independent of the distance $r$)
  • The electric potential at infinity(or, at infinite distance from the plate) is $0$.

Now we can easily work out your questions.

Does this mean there would be $0$ charge on their inner surfaces, leaving net $q$ and $kq$ for the outer ones respectively?

Yes, you are correct in saying that the charges will repel, but there will not be $0$ charge on their inner surfaces. It is because then the electric field inside the plates will no longer be $0$. The distance between the plates does not matter as the electric field of the plates is independent of distance. So now let's analyse the electric field at any one point inside the plates. If we want to make the field at that point zero, then the field due to charges on its left should be equal to the field due to the charges on its left. And since the field only depends on the charge density, which in turn only depends on the charges on the plates(as the areas are equal), so to have a zero electric field,

$$\sum q_{left} = \sum q_{right}$$

Now use this condition for the points inside the plates. You will get the charge distribution.

...the charges must be evenly split on the inner and outer surfaces of both plates otherwise the residue charges will cause a field in between the supposedly equipotential surfaces.

No. If you find the charge distribution using the above mentioned method, you will find that the charge on the inner surface of the plates is $0$. And all the charge reside on the outer surface of the plates. You can feel this intuitively by using the argument that the charges will repel each other and try to get as away as possible from each other. Moreover, here the electric field will be $0$ even if the all the charges run off to the outer surface. So there's nothing stopping them from doing so.

In the case where the plate is grounded, we first need to define the potential due to a charged infinite plane with respect to infinity. So,

$$\begin{align} V(r)-V(\infty)=V(r)&=-\int_{\infty}^r \mathbf{E}_{net}\cdot d\mathbf{r}\\ &=-\int_{\infty}^{r} \frac{\sigma_{net}}{2\epsilon_0}dr\\ &=-\frac{\sigma_{net}}{2\epsilon_0}r \Biggr|_{\infty}^{r}\\ &=\frac{\sigma_{net}}{2\epsilon_0}(\infty-r) \end{align}$$

But we want $V(r)=0$, which is only possible if $r=\infty$ or $\sigma_{net}=0$. And clearly, $r$ cannot be $\infty$, so $\sigma_{net}$ must be $0$. Hence the sum of the charges on all the plates must be $0$. So now you change the charge of the grounded plate such that the total charge becomes $0$. And now it is converted into a normal problem which can be solved using the method mentioned before in the answer.

  • $\begingroup$ Thanks for the good answer. $\endgroup$ Commented Dec 26, 2019 at 15:17

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