Earthed plates confusion

[tellmesomething]

Homework Statement

4 conducting plates with charges Q, -2Q, 4Q and 3Q initially are present as shown in the image attached. The third plate is then grounded later. Find the final charge at each surface of the plate

Relevant Equations

None

So, we know that since the 3rd plate is grounded that means the potential from the right side of the third plate to infinity should be 0, i.e no work should be required to bring a test charge from infinity to the right side of plate 3.

Similarly the left side of plate 3's potential should also be 0, i.e no work is done when moving a test charge from infinity to the left side of plate 3.

This should mean the fields in these regions should either cancel out or be 0.

My teacher told me that we should always start with the assumption that the two outermost surfaces of the system carry 0 charge and then conserve charge on each plate simultaneously till we reach the grounded plate.
Here when I do that i can see that the field in the left side of plate 3 , from infinity to Plate 1 is 0 therefore no work is done i.e potential is 0, and the field between plate 1 and plate 2 is opposite in direction to the field between plate 2 and plate 3, hence when we move from plate 1 to plate 3 the work done is 0.


On the right side of plate 3 work done by a test charge to move from infinity to plate 4 is zero since no field is there to move against but to move from plate 4 to plate 3 work done is non zero since theres a field, this work done is not even cancelled out by anything else so im confused how this can be non zero.

 

[kuruman]

Here when I do that i can see that the field in the left side of plate 3 , from infinity to Plate 1 is 0 . . .

Why? This means that the electric field on the left side of the plate array must be zero. If you construct a Gaussian pillbox that includes all four plates (see below), you see that the net charge enclosed is +6Q which, according to Gauss's law, means that there is a net electric field extending outward to infinity on both sides of the array. Your diagram shows that the total charge on all four plates is zero which contradicts the statement of the problem.

 

[tellmesomething]

Why? This means that the electric field on the left side of the plate array must be zero. If you construct a Gaussian pillbox that includes all four plates (see below), you see that the net charge enclosed is +6Q which, according to Gauss's law, means that there is a net electric field extending outward to infinity on both sides of the distribution. You diagram shows that the total charge on all four plates is zero which contradicts the statement of the problem.

View attachment 367385

This is the total charge on the plates initially before the third plate gets earthed. We have to find the charge distribution on the 4 plates after the third plate is earthed. The field you're talking about is before its earthed.

 

[kuruman]

This is the total charge on the plates initially before the third plate gets earthed. We have to find the charge distribution on the 4 plates after the third plate is earthed. The field you're talking about is before its earthed.

Then the statement of the problem "4 conducting plates with charges Q, -2Q, 4Q and 3Q are present as shown in the image attached. The third plate is grounded." is misleading. The attached image shows plate 3 already grounded and the total charge on each plate at that time. Is it possible that you misinterpreted or reinterpreted what your teacher gave you?

 

[Steve4Physics]

This is the total charge on the plates initially before the third plate gets earthed. We have to find the charge distribution on the 4 plates after the third plate is earthed. The field you're talking about is before its earthed.

The Post #1 diagram is misleading. It would be better to include an open switch in the earthwire and ask something like: "What are the final charges on each surface of the plate after the switch is closed?".

Edit. Aha. @kuruman beat me to it..

 

[kuruman]

The Post #1 diagram is misleading. It would be better to include an open switch in the earthwire and ask something like: "What are the final charges on each surface of the plate after the switch is closed?".

This reformulation makes sense.

 

[SredniVashtar]

The net charge of the system is not zero both before and after grounding the third plate. So I don't think you can assume the charge on the outermost surface has to be zero.

 

[Steve4Physics]

The net charge of the system is not zero both before and after grounding the third plate. So I don't think you can assume the charge on the outermost surface has to be zero.

The initial net charge is non-zero. But when plate-3 is grounded, charge will flow between the system and ground. The final net charge will be zero and the outer surfaces will have zero charge.

I think this result is quite general for conducting/earthed systems – providing there is no external electric field.

I'm now uncertain if what I said is correct. So I've struck it through. Apologies for any confusion caused.

 

[tellmesomething]

Then the statement of the problem "4 conducting plates with charges Q, -2Q, 4Q and 3Q are present as shown in the image attached. The third plate is grounded." is misleading. The attached image shows plate 3 already grounded and the total charge on each plate at that time. Is it possible that you misinterpreted or reinterpreted what your teacher gave you?

Oh right sorry that is probably the case, i copied it in class and the teacher did not give a proper question statement so i paraphrased, but yes this is what is implied, the plate 3 is grounded after.

 

[tellmesomething]

The initial net charge is non-zero. But when plate-3 is grounded, charge will flow between the system and ground. The final net charge will be zero and the outer surfaces will have zero charge.

I think this result is quite general for conducting/earthed systems – providing there is no external electric field.

Yes but the field is not zero to the immediate right of plate 3, if we move from plate 4 to plate 3 we are doing work against a field, and that means there's a non zero potential, but this shouldn't be the case...
this is what i was trying to ask in the post as well...

 

[tellmesomething]

The initial net charge is non-zero. But when plate-3 is grounded, charge will flow between the system and ground. The final net charge will be zero and the outer surfaces will have zero charge.

I think this result is quite general for conducting/earthed systems – providing there is no external electric field.

Like since at infinity potential is assumed to be 0, when we move from infinity (on either side) to plates A or D no work is done as these surfaces have 0 charge and there's no field to work against. When we move from plate A to C the potential difference also comes out to be 0 which is consistent with the fact that plate C is earthed. But when we move from plate D to C, theres a field which we have to work against and a non zero work is done, but since plate C is earthed this work done should also be 0, no?

 

[Steve4Physics]

View attachment 367391
Like since at infinity potential is assumed to be 0, when we move from infinity (on either side) to plates A or D no work is done as these surfaces have 0 charge and there's no field to work against. When we move from plate A to C the potential difference also comes out to be 0 which is consistent with the fact that plate C is earthed. But when we move from plate D to C, theres a field which we have to work against and a non zero work is done, but since plate C is earthed this work done should also be 0, no?

Hi @tellmesomething. I understand what you mean - sorry for not picking it up sooner. I’ve edited what I said in Post #8 as it may be incorrect/misleading.

I too had assumed the charges were distributed as you show in your Post #11 diagram. Maybe this is the problem. Or maybe there’s no requirement for the outer surface potentials to be zero (or even equal).

I’ll reflect and might post again (though someone else might sort it out in the meantime).

Minor edit.

 

[kuruman]

Like since at infinity potential is assumed to be 0, when we move from infinity (on either side) to plates A or D no work is done as these surfaces have 0 charge and there's no field to work against. When we move from plate A to C the potential difference also comes out to be 0 which is consistent with the fact that plate C is earthed. But when we move from plate D to C, theres a field which we have to work against and a non zero work is done, but since plate C is earthed this work done should also be 0, no?

Look at the figure on the right. It shows plates 3 and 4 and the standard pillbox Gaussian surface shown as a dotted line. There are surface charge densities as shown. Using Gauss's law is a better idea than using work done because it relates the surface charge density directly to the electric field.

Please answer the following questions
(a) What is electric flux through the left face of the pillbox which is entirely inside the earthed conductor?
(b) What is electric flux through the left right face of the pillbox which entirely to the right of the plates?
(c) What equation can you write that relates $\sigma_1$, $\sigma_2$ and $\sigma_3$?
(d) If the area of each face is $A$, you know that $\sigma_2~A+\sigma_3~A=3Q.$ In view of your answer in (c), what should $\sigma_1$ be equal to?

Use the same ideas to find the charge on the left side of the earthed conductor.

 

[SredniVashtar]

The initial net charge is non-zero. But when plate-3 is grounded, charge will flow between the system and ground. The final net charge will be zero and the outer surfaces will have zero charge.

I think this result is quite general for conducting/earthed systems – providing there is no external electric field.

I'm now uncertain if what I said is correct. So I've struck it through. Apologies for any confusion caused.

I believe it not to be true in general. It might be true with the unphysical case of infinite plates but the amount of charge exchanged with earth should in general be geometry dependent.

I asked an artificial helper to solve a problem with three charged spheres, the middle one of which to be earthed, and - unless it's allucinating - the total net charge after earthing is not zero.

 

[haruspex]

View attachment 367394Look at the figure on the right. It shows plates 3 and 4 and the standard pillbox Gaussian surface shown as a dotted line. There are surface charge densities as shown. Using Gauss's law is a better idea than using work done because it relates the surface charge density directly to the electric field.

Please answer the following questions
(a) What is electric flux through the left face of the pillbox which is entirely inside the earthed conductor?
(b) What is electric flux through the left right face of the pillbox which entirely to the right of the plates?
(c) What equation can you write that relates $\sigma_1$, $\sigma_2$ and $\sigma_3$?
(d) If the area of each face is $A$, you know that $\sigma_2~A+\sigma_3~A=3Q.$ In view of your answer in (c), what should $\sigma_1$ be equal to?

Use the same ideas to find the charge on the left side of the earthed conductor.

Doesn’t that result in the solution in post #11?
But I am uncomfortable with this question. Infinite plates and potentials is an awkward combination.
The standard approximation with infinite charged plates is that they generate a uniform field. Immediately we must drop the convention that the potential at infinity is zero. Instead, the potential at the plate itself is taken as the zero reference. That gives us no way to consider the potentials created by parallel charged sheets.
Suppose we take the potential at an infinite sheet of charge of density $\lambda$ to be $c\lambda$, for some unknown universal constant $c$. Placing a grounded sheet in parallel at distance $d$, if the induced charge density is $\mu$ then we can write:
$c\lambda-d\frac{\lambda}{2\epsilon_0}+c\mu=0$
which only yields $\mu=-\lambda$ in the limit $c/d\rightarrow \infty$.

Edit:
A possible resolution…
Consider two parallel discs radius R distance z apart, R>>z.
Disc A carries charge density $\lambda$, disc B is earthed and so acquires charge density $\mu$.
Of course, the charges won’t be uniform and neither will the fields, but glossing over that…
The potential at the centre of B due to A is (omitting a constant multiplier) $\lambda(\sqrt{R^2+z^2}-z)$, while that at the centre of B due to its own charge is $\mu R$. Hence $\mu=-\lambda(\sqrt{1+(\frac zR)^2}-\frac zR)$.
In the limit as $z/R\rightarrow 0$, $\mu=-\lambda$.

 

[haruspex]

I believe it not to be true in general. It might be true with the unphysical case of infinite plates but the amount of charge exchanged with earth should in general be geometry dependent.

Indeed, a standard example being the charge a point charge induces on a grounded sphere.

 

[kuruman]

Doesn’t that result in the solution in post #11?

Yes, however the connection between the explanation in post #11 and the charges shown in the figure is not, in my opinion, made rigorously. I am wondering whether the OP will be able to use his approach and arrive at the correct solution if the initial charges on the plates, before earthing, are given as $~Q_A$, $Q_B$, $Q_C$ and $Q_D.$

 

[tellmesomething]

View attachment 367394Look at the figure on the right. It shows plates 3 and 4 and the standard pillbox Gaussian surface shown as a dotted line. There are surface charge densities as shown. Using Gauss's law is a better idea than using work done because it relates the surface charge density directly to the electric field.

Please answer the following questions
(a) What is electric flux through the left face of the pillbox which is entirely inside the earthed conductor?
(b) What is electric flux through the left right face of the pillbox which entirely to the right of the plates?
(c) What equation can you write that relates $\sigma_1$, $\sigma_2$ and $\sigma_3$?
(d) If the area of each face is $A$, you know that $\sigma_2~A+\sigma_3~A=3Q.$ In view of your answer in (c), what should $\sigma_1$ be equal to?

Use the same ideas to find the charge on the left side of the earthed conductor.

well yes a) 0 b) $\frac{\sigma_1 A}{2\epsilon}+ \frac{\sigma_2 A}{2\epsilon} +\frac{\sigma_3 A}{2\epsilon}$ (A here being the area of the pillbox) c) $\sigma_1 + \sigma_2 + \sigma_3=0$ d) $\sigma_1= \frac{-3Q}{A}$ (A here being the area of the infinite plates)

 

[kuruman]

A possible resolution…

Here is my resolution. Consider two concentric conducting shells. Charge $Q$ is placed on the inner shell and then the outer shell is earthed. The inner shell has outer radius $R$ and the gap between shells is $d$. What are the surface charge distributions on the shells? (See figure on the right.)

Answer
Construct a concentric spherical Gaussian surface entirely within the outer shell. The electric field is zero everywhere on this surface, hence $\int_S \mathbf E\cdot \mathbf{\hat n}~dA=0.$ This means that the total charge enclosed by the Gaussian surface is zero. We already know that there is charge $Q$ on the inner shell, therefore charge $-Q$ must flow from the Earth to the inner surface of the outer conductor when it's earthed.

All the charge on the inner shell resides on its outer surface. By symmetry, the charge on the surface of each conductor is uniform. Thus, $$\sigma_1=\frac{Q}{4\pi R^2}~;~~\sigma_2=-\frac{Q}{4\pi(R+d)^2}.$$ Now imagine the inner shell being the size of the Earth and $d$ of order of a few centimeters. Since $R>>>d,$ it follows that $\sigma_2\approx -\sigma_1.$

A human standing on the inner conductor constructs a Gaussian pillbox as indicated by the dashed lines in the figure (not drawn to scale). To this human, the pieces of conductors enclosed by the Gaussian surface appear parallel whilst the conductors themselves recede to infinity.

I note that both resolutions require some kind of approximation argument to deal with the idea of flat, infinite plates.

(Edited for typos)

 

[Steve4Physics]

@tellmesomething, this might resolve specific issues you raised….

So, we know that since the 3rd plate is grounded that means the potential from the right side of the third plate to infinity should be 0, i.e no work should be required to bring a test charge from infinity to the right side of plate 3.

‘Ground’ and ‘infinity’ are not necessarily equivalent. I believe that we can only say that no work is done bringing a test charge from ground to the right side (or indeed to the left side) of plate 3.

Yes but the field is not zero to the immediate right of plate 3, if we move from plate 4 to plate 3 we are doing work against a field,

Minor (pedantic) point - it might be better to say: if we move a positive test charge from plate 3 to plate 4, then we are doing [positive] work against the field.

and that means there's a non zero potential, but this shouldn't be the case...

The potential at the right surface of face 3 is zero (ground potential). There is a potential gradient between plates 3 and 4. ($\vec E = -\frac {dV}{dx} \hat i$.) The potential increases from 0 to some positive value as we move from plate 3 to plate 4.

Think of a simplified version of the system as a charged capacitor with one plate grounded:

On a more general point, we are assuming that the linear dimensions of each plate are very large compared to the plate-separation. This is not explicitly stated in the question. It means that edge effects can be ignored and we can make the simplifying assumption that opposing faces of adjacent plates have equal and opposite charges.

 

[haruspex]

both resolutions require some kind of approximation argument to deal with the idea of flat, infinite plates.

But I don’t see a valid argument for that in the concentric sphere model. A single infinite charged plate clearly exerts a field on both sides, but if we model it as the limit of part of a spherical surface then, on the side taken to be the interior, there is no field.

 

[haruspex]

we are assuming that the linear dimensions of each plate are very large compared to the plate-separation. This is not explicitly stated in the question. It means that edge effects can be ignored and we can make the simplifying assumption that opposing faces of adjacent plates have equal and opposite charges.

This is what I am missing. How does that follow?

 

[kuruman]

A single infinite charged plate clearly exerts a field on both sides, . . .

Why? If the plate is truly infinite, it must be at the reference value of the potential at infinity, i.e. earthed, at all times. If you place charge Q on one side, it will leak out to infinity. I think that my concentric shell picture (I wouldn't call it a model) gets around that problem and gives the correct surface charge distributions.

 

[haruspex]

Why?

It is the model everyone uses, e.g. https://phys.libretexts.org/Bookshe...d_of_a_Uniformly_Charged_Infinite_Plane_Sheet

Of course, infinite sheets don't exist, so what really is the charge distribution on a very large sheet?
Figure 5 at https://www.researchgate.net/public...as_the_Application_of_Far-Ranging_Mathematics
suggests that, in the limit (but presumably keeping the total charge as proportional to the area), the charge does not tend to zero at the centre. But I have not yet found there an actual proof of that.
Not sure I trust this paper, but the equations are impressive and I've not found any others.

Another difficulty is that an infinite uniform density would generate an infinite potential and infinite field, right?

 

[Steve4Physics]

This is what I am missing. How does that follow?

Well, I'm not sure if this answers your question but here’s a qualitative (definitely not rigorous) attempt...

Thinking visually, a field line starts on a positive charge and ends on a negative charge.

Consider two (finite) parallel conducting plates arranged as shown below. The upper plate is positively charged and the lower plate is grounded. The resulting charges and field will be like this:

Considering the symmetry and the fact that the field inside the conductors must be zero, we can justify (using Gauss’s law) that the induced charges on the top surface of the lower plate must match the charges on the lower surface of the upper plate. This is true even when edge-effects are present because of the symmetry.

(Note, this is independent of the surface charge density distribution – we are only interested net surface charges.)

Now consider what can happen if the symmetry is broken. A simple example is introduction of an additional point charge:

The induced charge on the lower plate is reduced.

For given plates, edge-effects can be reduced by having a smaller gap between plates, essentially because the amount of flux that can pass through the gap's edge is restricted.
_________

Irrelevant note. What new word have I learned today? It’s “indagation” (from Post #24 link https://www.researchgate.net/public...as_the_Application_of_Far-Ranging_Mathematics).

 

[haruspex]

Consider two (finite) parallel conducting plates

Sure, but that’s just two plates. What if, e.g., we start with two positively charged plates then add a grounded plate, not between them? Even if that works, how to show it in general?

 

[Steve4Physics]

Sure, but that’s just two plates. What if, e.g., we start with two positively charged plates then add a grounded plate, not between them? Even if that works, how to show it in general?

I may misunderstand the issue you are indigating. But to try to answer your question...

Take three conducting plates (equal sizes, aligned, parallel, negligible gap compared to linear dimensions). One outer plate is grounded and the other two plates have charges of, say, +5Q and +2Q. I believe the surface charges will distributed like this:

There are infinitely many other possibilities such as this:

However, the outer face of the grounded plate is uncharged (or charge would flow to /from ground to neutralise it). I believe this requires that the other outer face (of the left plate) must also be uncharged*. This corresponds to the first solution.

*E.g. see this: https://physics.stackexchange.com/q...ies-on-a-system-of-parallel-charged-plates-id

 

[kuruman]

There are infinitely many other possibilities such as this: . . .

I am not so sure about that. The charge distributions on all surfaces can be sorted out by using Gaussian pillboxes. Refer to the figure on the right.

1. Construct Gaussian pillbox 1 enclosing all three plates (black). The total enclosed charge must be zero to eliminate the electrostatic potential difference between the volume enclosed by pillbox 1 and the Earth. As long as this difference is non-zero, there will be electric field lines that will move charges to/from the Earth from/to the earthed conductor. By Gauss's law, the charge on the earthed plate must be the negative of the sum of the charges on the other two plates, $Q_C=-(Q_A+Q_B).$

2. Now construct Gaussian pillboxes 2A and 2B (red). They have one face entirely inside a conductor and another face outside pillbox 1 where the field is zero. This means that the total electric flux through pillboxes 2A and 2B is zero. Hence, they each enclose zero charge, i.e. the sides of the plate assembly facing the outside have zero charge on them. This is the case only because one of the plates, it doesn't matter which, is earthed.

3. Finally construct pillbox 3 (blue). As with pillbox 2, the charge enclosed by pillbox 3 must be zero. This means that the surface charge densities on the sides of plates C and B facing each other are equal and opposite. Likewise for the sides of plates B and A facing each other.

The rest is algebra.

N.B. If the plates are finite, I think that the arguments above will still be valid as long as pillbox 1 is large enough to enclose all three finite plates and the charge on these plates is all there is.