# Conducting slab between the plates of parallel plate capacitor

1. Jun 26, 2013

### Tanya Sharma

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

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

2. Relevant equations

3. 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 ?

2. Jun 26, 2013

### Staff: Mentor

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.

3. Jun 26, 2013

### sankalpmittal

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.

4. Jun 26, 2013

### Tanya Sharma

Hi gneill

Thanks for the response

What you have explained sounds logical.Initially I too,thought on the similar lines,but the first explaination ,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=∞ ?

5. Jun 26, 2013

### Staff: Mentor

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.

6. Jun 26, 2013

### Tanya Sharma

Nice explaination...

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 ?

7. Jun 26, 2013

### Staff: Mentor

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.

8. Jun 26, 2013

### Tanya Sharma

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?

9. Jun 26, 2013

### Staff: Mentor

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

10. Jun 26, 2013

### Tanya Sharma

Thanks gneill

11. May 5, 2016

### 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 ??

12. May 5, 2016

### Staff: Mentor

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.

13. May 5, 2016

### Surya7

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

14. May 5, 2016

### Staff: Mentor

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.

15. May 5, 2016

### 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 ??

16. May 5, 2016

### 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

17. May 5, 2016

### Staff: Mentor

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.

18. May 6, 2016

### 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).

19. May 6, 2016

### cnh1995

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.

20. May 6, 2016

### conscience

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 .

Last edited: May 6, 2016