Conducting slab between the plates of parallel plate capacitor

[Tanya Sharma]

Homework Statement

The space between the plates of a parallel plate capacitor is completely filled with a conducting slab.What is the capacitance of the system ?

Homework Equations

The Attempt at a Solution

The capacitance of a capacitor when a dielectric medium is placed between the plates is given by :

C=εoA/(d-t+t/K)

Where, εo is permittivity of free space, A is the area of a plate , d is the distance between the plates , t is thickness of the dielectric medium, K is dielectric constant of the medium.

Now if the conductor completely occupies the space, then K=∞ and t=d

Hence we get, C=∞ .

Now look at the same situation in terms of the definition of capacitance i.e Q=CV .

The capacitance is the ability of the plates to hold charge. But here if we connect the plates of the capacitor to the terminals of the battery , the current flows through the conducting slab ,across the plates .No charge will be accumulated on the plates .

The potential difference across the plates is zero.

What should we put as Q in Q=CV ? If we put zero ,as is the case, then we have C = 0/0 ,which is undefined .

What is the correct way of looking at the problem ?

 

[gneill]

When you bridge the gap between the plates with a conductor you no longer have a capacitor. Trying to determine the capacitance of something that isn't a capacitor is a logical trap

Your analysis is equivalent to taking a capacitor without dielectric and determining the capacitance as the plates are brought together (the distance d is allowed to approach zero). Sure, the capacitance gets bigger and bigger as the separation diminishes. But once the plates touch, you've no longer got a capacitor because charge will not remain separated in the device.

 

[sankalpmittal]

Homework Statement

The space between the plates of a parallel plate capacitor is completely filled with a conducting slab.What is the capacitance of the system ?

Homework Equations

The Attempt at a Solution

The capacitance of a capacitor when a dielectric medium is placed between the plates is given by :

C=εoA/(d-t+t/K)

Where, εo is permittivity of free space, A is the area of a plate , d is the distance between the plates , t is thickness of the dielectric medium, K is dielectric constant of the medium.

Now if the conductor completely occupies the space, then K=∞ and t=d

Hence we get, C=∞ .

Now look at the same situation in terms of the definition of capacitance i.e Q=CV .

The capacitance is the ability of the plates to hold charge. But here if we connect the plates of the capacitor to the terminals of the battery , the current flows through the conducting slab ,across the plates .No charge will be accumulated on the plates .

The potential difference across the plates is zero.

What should we put as Q in Q=CV ? If we put zero ,as is the case, then we have C = 0/0 ,which is undefined .

What is the correct way of looking at the problem ?

Hello there,

Its a homework question, so I will just guide you.

When a conductor is placed between the parallel plate capacitor, then what will be the E-field inside the conductor ? Also, what will be its dielectric constant then ?

Now,

Imagine, when you let the charge to flow "ONTO" the conductor, THEN no charge will be accumulated on it. Hence by definition (capacitance is ability of conductor to HOLD charge) capacitance will be zero.

For example, a simple wire. Its not a capacitor. But theoretically its capacitance is zero, by definition. Although we do not use Q=CV there because Q is charge accumulated remember, and is only applicable when a body is a capacitor.

Now imagine you have a parallel plate capacitor. You put a conductor there and fill the whole thickness. Already the polarities are created and hence body will be CAPACITOR. Hence in this case you can use formulas of capacitance. One such formula is C=εoA/(d-t+t/K). Putting t=d and k=∞, you get C=∞..

Now the third case :

Suppose the capacitor is short, that is potential difference across it is somehow zero. Practically no charge will get accumulated on it. Hence using C=Q/V, we get C=0/0... Hence its capacitance can be anything. But we merely neglect it while finding effective capacitance of a circuit. Note that its still a capacitor.

 

[Tanya Sharma]

When you bridge the gap between the plates with a conductor you no longer have a capacitor. Trying to determine the capacitance of something that isn't a capacitor is a logical trap

Your analysis is equivalent to taking a capacitor without dielectric and determining the capacitance as the plates are brought together (the distance d is allowed to approach zero). Sure, the capacitance gets bigger and bigger as the separation diminishes. But once the plates touch, you've no longer got a capacitor because charge will not remain separated in the device.

Hi gneill

Thanks for the response

What you have explained sounds logical.Initially I too,thought on the similar lines,but the first explanation ,i.e C=∞ has been given in a couple of books .This forced me to think somewhat differently.

Is it possible that the question be framed somewhat differently to get C=∞ ?

 

[gneill]

Hi gneill

Thanks for the response

What you have explained sounds logical.Initially I too,thought on the similar lines,but the first explanation ,i.e C=∞ has been given in a couple of books .This forced me to think somewhat differently.

Is it possible that the question be framed somewhat differently to get C=∞ ?

Well, you might say that in the limit as the gap approaches zero the capacitance tends to infinity. But it's a one-sided limit, as the capacitance is undefined when d = 0 (or d < 0).

In any practical device the breakdown voltage of the gap will limit just how small the gap can be for a given working voltage.

 

[Tanya Sharma]

Well, you might say that in the limit as the gap approaches zero the capacitance tends to infinity. But it's a one-sided limit, as the capacitance is undefined when d = 0 (or d < 0).

In any practical device the breakdown voltage of the gap will limit just how small the gap can be for a given working voltage.

Nice explanation...

What if ,we were able to hold the charges on the plates .Say,we insert two infinitesimal width insulating sheets between the plates and the conductor .Is capacitance infinite in this case ?

 

[gneill]

Nice explanation...

What if ,we were able to hold the charges on the plates .Say,we insert two infinitesimal width insulating sheets between the plates and the conductor .Is capacitance infinite in this case ?

Depends on what you mean by "infinitesimal" and whether we're talking theory or practice. Actual infinite capacitance demands that Δx → 0 in the calculus sense. No real material can do that. Otherwise the capacitance will just be very large. Thin film capacitors like electrolytics employ a very thin film (a few atoms or molecules thick) as an insulator. Bigger capacitors with thinner films tend to have lower working voltages.

On the practical side where things are made of real materials comprised of atoms, surfaces are not smooth and distances become hard to define, and vary dynamically and randomly due to thermal activity of the atoms. Quantum effects will eventually squash any attempts to improve matters as the uncertainty principle sees to it that charges will be able to tunnel across any intervening gap, even if there's a "perfect" insulator in the way.

 

[Tanya Sharma]

Depends on what you mean by "infinitesimal" and whether we're talking theory or practice.

Well, I meant in theory .

Isnt filling the complete space between the plates with a conducting slab same as shorting the capacitor ?

I mean,if we join the plates with a conducting wire ,or, a thin metal plate is inserted between the plates in such a way that its edges touch the two plates,then the system of parallel plates no longer qualifies itself as capacitor .Right?

 

[gneill]

Well, I meant in theory .

Isnt filling the complete space between the plates with a conducting slab same as shorting the capacitor ?

I mean,if we join the plates with a conducting wire ,or, a thin metal plate is inserted between the plates in such a way that its edges touch the two plates,then the system of parallel plates no longer qualifies itself as capacitor .Right?

Right. That's what I stated in post #2.

 

[Tanya Sharma]

Thanks gneill

 

[Surya7]

In general what will be used to fill the gap between the conductors in a capacitor , is it completely insulating medium or a conducting Slab and why ??

 

[gneill]

In general what will be used to fill the gap between the conductors in a capacitor , is it completely insulating medium or a conducting Slab and why ??

It is an insulator. Although there are no perfect insulators made of real materials, the purpose of the dielectric is to modify the electric field between the plates, not to conduct charges across the gap.

 

[Surya7]

But when we calculate the capacitance by placing a conducting slab is more than that of when we use a dielectric( insulating medium) ?

 

[gneill]

But when we calculate the capacitance by placing a conducting slab is more than that of when we use a dielectric( insulating medium) ?

If it's conducting then it is not a dielectric --- it forms another plate. You then end up, effectively, with two capacitors in series, each with much smaller gaps than the original capacitor since the middle "plate" takes up space in the original gap.

 

[Surya7]

Yes , then series of two capacitors should lead to lesser capacitence, but in other logic, when ever a conducting slab placed in between two fields net electric field is zero across slab ( in presence of electric field ) that leads to lesser potential means. More capacitence , but when we have insulating slab there net electric field is not zero , so potential difference increases means lesser capacitence ? Please correct me if I am wrong ??

 

[Surya7]

Capacitence when I place a conducting slab
C= εoA/(d-t)
Capacitence when I place insulating medium
C=εoA/(d-t+t/K)
From above equations , I feel capacitence is more in first case... I derived this equations using the logic from my previous post

Yes , then series of two capacitors should lead to lesser capacitence, but in other logic, when ever a conducting slab placed in between two fields net electric field is zero across slab ( in presence of electric field ) that leads to lesser potential means. More capacitence , but when we have insulating slab there net electric field is not zero , so potential difference increases means lesser capacitence ? Please correct me if I am wrong

 

[gneill]

Yes , then series of two capacitors should lead to lesser capacitence, but in other logic, when ever a conducting slab placed in between two fields net electric field is zero across slab ( in presence of electric field ) that leads to lesser potential means. More capacitence , but when we have insulating slab there net electric field is not zero , so potential difference increases means lesser capacitence ? Please correct me if I am wrong ??

Capacitance increases with a dielectric, for the same total gap. Look at the basic formula for capacitance of a parallel plate capacitor:

##C =k \epsilon_o \frac{A}{d}##

where ##\epsilon_o## is the permittivity of the vacuum and k is the relative permittivity of the dielectric medium. So the effective permittivity of the region between the plates with a dielectric in it is ##k\epsilon_o##. We assume here that the dielectric constant k is greater than unity, which it is for materials used as dielectrics.

The main reason that dielectrics are used to increase capacitance rather than simply reducing the spacing between the plates is that the highest voltage that can be placed between the plates (breakdown voltage) is inversely proportional to the gap size. At a critical strength of electric field breakdown will occur, even for hard vacuum. The field is measured in V/m, so the critical field strength is reached sooner with smaller gaps.

 

[Surya7]

Superb explanation , still I have more doubt I.e what substance( assume it has thickeness t) will be used in between the plate so that we arrive at this expression for capacitence.
Capacitence
C= εoA/(d-t).

 

[cnh1995]

what substance will be used in between the plate so that we arrive at this expression for capacitence.

I believe it should be a metallic slab of thickness t. If d=t(i.e. the dielectric is replaced by a conducting slab), the expression gives an infinite capacitance, which is true for conductors. Conductors have infinite permittivity.

 

[conscience]

If d=t(i.e. the dielectric is replaced by a conducting slab), the expression gives an infinite capacitance, which is true for conductors.

If d = t , the plates do not form a capacitor anymore . If the plates cannot hold charge then what is the significance of saying that the capacitance is infinite .This is exactly what @gneill has clarified that it is not right to say that plates have infinite capacitance when , either the space between plates is completely filled by a conductor , or the plates are shorted( i.e connected by a wire) .Instead it would be correct to say that capacitance is undefined. Please read post#2 and #5 .

 

[cnh1995]

If d = t , the plates do not form a capacitor anymore

It is not a capacitor practically, true. But I believe mathematically the capacitance can be said to be infinite. It will take an infinite amount of charge to establish an electric field inside the conductor, which practically means no charge can be stored (since it is conducting). I believe mathematical infinite capacitance means it is not a capacitor practically..

 

[cnh1995]

Conductors have infinite permittivity.

If you apply an external electric field E to a dielectric with a dielectric constant k, the dielectric will be polarized and there will be an opposing electric field in the dielectric. So the net field in the dielectric is E^'=E_original-E_polarization. Relative permittivity is defined as "ratio of applied electric field to the net electric field", i.e. k=E_original/E'. Now in case of conductors, net electric field E' will always be 0. This gives k=∞ (or permittivity of conductors is infinite).

 

[conscience]

But I believe mathematically the capacitance can be said to be infinite. It will take an infinite amount of charge to establish an electric field inside the conductor, which practically means no charge can be stored (since it is conducting).

I disagree . No matter how much charge is present on the plates,electric field is zero inside the conducting material .This doesn't imply an infinite charge is required . Also , infinite permittivity of conductors is not a justification that when d = t , capacitance is infinite . Capacitance is undefined in this case .

 

[Surya7]

I believe it should be a metallic slab of thickness t. If d=t(i.e. the dielectric is replaced by a conducting slab), the expression gives an infinite capacitance, which is true for conductors. Conductors have infinite permittivity.

C=eoA/(d-t+t/K) -- Effective capacitence when we use a Di electric material (insulator) of thickeness t
C=eoA/(d-t) -- Effective capacitence when we use a metalic slab of thickeness t
From the above equations it is evident that it is better to use Metalic slab instead of Insulator, since it has More Capacitance (more ability to store charge) ? But why all the practical capacitors are filled by dielectric (insulator) ?

 

[gneill]

But why all the practical capacitors are filled by dielectric (insulator) ?

This was answered in posts #7 and #17. The bottom line is that there is a practical lower limit on how small the gap can be without the capacitor failing due to arc-over.

 

[cnh1995]

From the above equations it is evident that it is better to use Metalic slab instead of Insulator,

If you look at the mechanism of charge storage, you can see it is achieved by polarization of the dielectric. Polarization implies no conduction and hence, charges are trapped on the plates. If the dielectric is replaced by a metallic slab, there will be conduction and no charge will be stored on the plates. In dielectrics, the opposing field due to polarization is less than the original applied field, hence you get a non-zero net (but reduced) electric field in the dielectric. To bring the net field in the dielectric back to the original applied magnitude, more charge is needed on the plates, which is provided by the source in the circuit. When the dielectric is replaced by a conducting slab, the net electric field becomes 0 i.e. conductor produces an opposing field equal to the applied field. This means, if you were to bring the net field back to the original applied magnitude, you'd need an infinite amount of charge, which means it will never happen practically. If such a capacitor is placed in a circuit, it will simply conduct and no charge will be stored. Hence, practically it is no longer a capacitor.