# Charge on metal plates

1. Jun 4, 2013

### Tanya Sharma

1. The problem statement, all variables and given/known data

When a charge Q is given to an isolated metal plate X of surface area A its surface charge density becomes σ .When an earthed metal plate Y is brought closed to X ,find magnitude of surface charged density on inner and outer surfaces of X and Y.

2. Relevant equations

3. The attempt at a solution

When charge Q is given to X it gets equally divided on the two surfaces of X i.e σ =Q/2A

If Y is not earthed then
When Y is brought close to X ,the charge induced on inner surface of Y will be equal and opposite to that of inner surface of X i.e -Q/2 and charge on outer surface of Y will be +Q/2 .

But Y is earthed in the given problem.Potential of Y will be zero .

I am not able to distribute charges .How earthing affects the distribution of charges ?

I would be thankful if some guidance is provided.

2. Jun 4, 2013

### ehild

The potential of the earthed plate is the same as in infinity. What can be the electric field between the earthed plate and infinity?

How is the electric field near a metal plate related to the surface charge density?

ehild

3. Jun 4, 2013

### rude man

Call the surfaces 1,2,3 and 4 from left to right. Assume the left plate is "earthed" and the right one carries charge Q.

1. Run a gaussian box from just to the right of surface 2 to just to the left of surface 3.
(or run it from inside plate 1 to inside of plate 2).
2. Run a gaussian box from inside plate 1 to just outside surface 4.
3. Use conservation of charge on both plates.

You get 4 equations and 4 unknowns & solve for all four surface charges.

EDIT: nuts, I only get 3 equations!

Last edited: Jun 4, 2013
4. Jun 4, 2013

### Tanya Sharma

Since the potential difference between the earthed plate and infinity is zero ,there can be no E field between the two.That means the outer surface will have no charge.

E = σ/2ε0 due to a single infinite charged surface

5. Jun 4, 2013

### ehild

Speaking about surface charge density, it can mean two things. The charge of unit cross section of a sheet, or the charge of unit area on a metal surface. You know that the electric field lines emerge from and end in charges, and a unit charge is source of 1/ε field lines. If the electric field is zero at one surface of a metal sheet, there are no surface charges on that surface. If the sheet is charged, the whole charge appears on the other surface.

In case of a single metal sheet of charge Q and area A, the charge distributes evenly on the whole surface. Ignoring the charge on the edges, the charge density on both surfaces is Q/(2A) because of symmetry. The electric field is also symmetric, E=Q/(2Aε0). It is not the case when you bring an earthed metal sheet close to the first one: the charged metal plate attracts opposite charges from the ground to the earthed plate and you get two oppositely charged plates at the end.

ehild

6. Jun 5, 2013

### Tanya Sharma

Thanks ehild

I have got the correct answer .I have to admit I was really struggling with earthing .I think I have now understood the concept.

If now we bring another metal plate Z with charge Q ,to the right of Y .i.e we have X Y Z (with Y earthed and equal spacing between X and Y and Y and Z) ,we will have +Q on inner surface of X,-Q on both the surfaces of middle plate Y ,and +Q on inner surface of Z .We essentially have two parallel capacitors connected in series .

Is my thinking right ?

7. Jun 5, 2013

### ehild

You can solve problems where the electric field of charged metal plates is the question that you substitute the plates with sheet of charges, like in the figure, and apply the principle of superposition. One plate will produce e=Q/(2Aε) field lines per unit area at both sides. In the first region in the figure, the net electric field is E(1)=-e1-e2. In the middle region, it is E(2)=e1-e2. In the third region, E(3)=e1+e2.

I case the right sheet has zero charge and Q1=Q, e1=Q/(2Aε) and e2=0. E(1)=-Q/(2Aε), E(2)=Q/(2Aε), E(3)=Q/(2Aε).

In case the right sheet is earthed, E(3)=0. What is its charge then?

ehild

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8. Jun 5, 2013

### ehild

That is right. The ground is infinite source of charges, and charges will flow to the grounded plate till the whole arrangement becomes electrically neutral.

Yes, it is equivalent to two capacitors, a pair of plates connected. But series and parallel has meaning only when you connect them into a circuit. If you connect a battery across the free terminals, they are in series. If you connect the other plates together and connect a battery between the earth and the other plates, the capacitors are parallel.

ehild

9. Jun 5, 2013

### Tanya Sharma

Okay...Suppose we have three metal plates,X,Y and Z .X earthed (neutral),Y (charge Q),Z(neutral) .In this case a single parallel plate capacitor is formed with X having -Q charge on inner surface and Y having +Q charge on inner surface .Z will have no role to play .

Is it correct ?

10. Jun 5, 2013

### ehild

It is right.

ehild

11. Jun 5, 2013

### Tanya Sharma

Thank you very much for the nice picture and beautiful explainations .All my doubts have been cleared.

You are a mentor .

12. Jun 5, 2013

### ehild

Thank you for the thanks, but I am not a person, who gives infractions and bans people. :tongue2:

ehild

13. Jun 5, 2013

### rude man

@ehild: " The ground is infinite source of charges, and charges will flow to the grounded plate till the whole arrangement becomes electrically neutral. "

I agree that the grounded plate gets a charge of -Q, but I was wondering what was your rationale for stating that. The statement that the necessary negative charges are sucked out of the ground is of course true but does it explain why? The argument that charges on the +Q plate must end on like charges on the grounded plate is also nebulous to me. Could those +Q charges not end on ground charges on the far side of the +Q plate, for example? Or even on induced negative charge on the grounded plate, the other side assuming an equal but opposite surface charge so that the overall grounded plate charge is zero instead of -Q?

I have come up with my own rationale for the fact that the grounded plate gets -Q of charge, but before submitting it to you for critique I wondered if you had another way of looking at this charge induction on top of what you've already said.

Thanks.
rude man

14. Jun 6, 2013

### ehild

Till equilibrium is established, there is electric field in the metals. Anyway, it consists of atoms and electrons and -mostly- of void. Assume one plate is charged positive, with charge Q. There is electric field around the charged plate, so attractive force is exerted on the electrons of the grounded plate when it is placed near to the first one. These electrons go as close as possible to the positive plate, they accumulate on the surface, opposite to the charged plate. In equilibrium, no force acts on the electrons in the bulk of the grounded metals. The electric fields due to the charged plate and the negative surface charges on the grounded plate have to cancel each other inside the metal plates.

The grounded plate is at the same potential as infinity. Because of planar symmetry, the electric field lines are normal to the plane of the plate and the electric field is equal to the negative gradient of the potential, which is zero. The electric field on the far side of the grounded plate is zero. At the surface, the electric field is equal to the surface charge density over ε. The surface charge density has to be zero on the far side.

You can fill the whole volume from the grounded plate to infinity with metal, it will not change the electric field on the other side: that means "the far side" of the grounded plate is infinite far. If you assume charges on the far side of the charged plate and field lines emerging from them and going to infinity, the work done along that infinite long path has to be equal to the work along the shortest path between the plates. That means zero electric field at the outside of the charged plate.
That means all the electric field lines emerging from the charged plate are between the plates and end in the induced surface charges on the inner surface. The number of the electric field lines is Q/ε. So the surface charge on the inner surface of the grounded plate is -Q.

First, when I met such problems of parallel charged metal plates, I also tried to solve them with Gauss' Law, but it was not enough. I learnt the method of superposition here, in the Forums. The method gives a solution of the Maxwell equations and obeys the boundary conditions, and I also learnt that such a solution is unique.

I am looking forward to your explanation.

ehild

Last edited: Jun 6, 2013
15. Jun 6, 2013

### rude man

Thanks for the reply. I will be mulling it over for some time I think.

OK, here is a circuits-oriented explanation: plate 1 is grounded and uncharged, plate 2 is placed facing plate 1 and initially uncharged also. So Q1 = Q2 = 0.

Now we run a +dc emf source V from ground to plate 2. Assuming ground conductivity is > 0 we have a closed circuit with the two plates forming a capacitor of value C. This capacitor will charge up via the emf and ground resistance to Q2 = CV. So plate 2 will assume Q2 = + CV and plate 1 will have Q1 = -CV, just as with all capacitors. The charging current added charge to plate 2 and removed charge from plate 1. We now remove the emf source so plate 2 floats. But that action changes nothing; opening one or both ends of any charged capacitor does not remove the charges on each plate. So we wind up with Q1 = - Q2.

However: since then I asked myself: what if plate 2 were charged to Q2 remotely and then brought face-to-face with plate 1? Now there is no circuit and no current to remove charge from plate 1. So plate 1 should remain uncharged.

Dilemma!

16. Jun 7, 2013

### ehild

Plate 2 has capacitance with respect to the ground. When it is charged to Q, you can imagine that the charge was brought from the ground. You have plate 2 and the ground (with charge -Q) as the other plate of capacitor C2. When the grounded plate P1 is brought close to P2, you have two capacitors connected through the earth, as in the picture on the right. But capacitor C2 is much - much smaller than C1, so the charge on the common terminals is shared so that (almost) the whole charge is on C1. So P1 acquires -Q charge when C2 can be ignored with respect to C1.

ehild

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