# Removal of dielectric from a charged capacitor.

1. Oct 6, 2011

### sodaboy7

There is a parallel plate capacitor having capacity C. It initially has got no charge on it. Now we insert a dielectric material of dielectric constant K between its plates (it still has no charge). Now we connect this capacitor (with dielectric) to a d.c source of potential difference V. Now charge builds up on this capacitor till its fully charged. WHAT WILL HAPPEN IF WE REMOVE THE DIELECTRIC WHILE THE CAPACITOR IS STILL CONNECTED TO V ??
According to my knowledge the dielectric increases the capacity of a capacitor by a factor of k i.e now it will a charge kq which is more than what it would had held (q) without the dielectric. So when we remove the dielectric its capacity to hold additional charge (kq-q) is lost. So what happens to this additional charge kq-q ???

2. Oct 6, 2011

### tsoits

Charges will flow back through the d.c. source so that the charges stored in the capacitor will decrease to q.

3. Oct 7, 2011

### Delta2

In order for charges to flow back, this means that a current reversed to the polarity of the d.c source will flow, which means that the voltage on the capacitor will be higher than that of the d.c source.. How will the voltage grow bigger???

4. Oct 7, 2011

### skmaidulhaque

I think, it is like this.
When there was dielectric witin the capacitor and the circuit was connected, there was a steady current, I in the circuit from +ve to -ve external to the DC source. Now, when you remove the dielectric, some amount of charge will flow in the direction as opposed to I. Thereby, it would reduce the current I by some amount. But, stilll current will flow in the same direction.

5. Oct 7, 2011

### Delta2

Thats not true. If the total(net) current keeps flowing in the same initial direction then the total charge on the capacitor will keep increasing. But we have a net decrease on the charge from kq down to q. This can be done only by a total reversed current.

6. Oct 7, 2011

### tsoits

Because of the absence of the dielectric!

7. Oct 7, 2011

### sodaboy7

We should consider following points.

1) We are forgetting that dielectric is closely related to electric field between the plates. It actually REDUCES the electric field by a factor of K. When dielectric is removed some work has to be done to compensate loss of electric field.

2) no current flows through the capacitor when it is fully charged

3) If current flows in opposite direction of convention the some heat must dissipate out.

4) Lets say we are removing the dielectric with our hands, then our hands must experience some force.

8. Oct 7, 2011

### tsoits

There're definitely energy exchange in the transient stage. But here we simply have to look at the steady state, and we can deduce what have happened.

As you mentioned in the first post, when removing the dielectric, the capacitance has decreased,
C = Q / V
V is constant, Q must have decreased.

When you consider the electric field,
E = σ /ε
when removing the dielectric, ε decreased.
However, don't forget that
E = V / d
with V and d are constant, so E must be constant as well.
So what have happened is that surface charge density σ has decreased too, which is resulted from Q stored decreased.

Everything is consistent, so during the transient period, the extra charges must have gone back to the d.c. source.

Please don't take me in the wrong way, but this is actually quite a standard problem in a lot of textbooks.

9. Oct 7, 2011

### sodaboy7

So if we add an ammeter in the same circuit and then remove the dielectric, the ammeter should show some current in a direction opposite to conventional polarity. Is it so??
Also is it possible that this charge may dissipate/leak out like electrical breakdown (or corona)??
Or is it possible that the movement of extra charges may cause some heat generation??

10. Oct 7, 2011

### tsoits

1) Yes. There will be an ammeter reading. Of course that depends on how sensitive the ammeter is. Using a galvanometer may be better.

2) You may regard it as a discharge process. So there will be joule heating at the internal resistance of the battery, the circuit, etc.
Note that the energy stored in the capacitor afterwards is smaller than before. Some energy also comes from the work done in removing the dielectric.

3) Electrical breakdown is unlikely. Unless the electric field is larger than the dielectric strength of air, making it ionized..

11. Oct 8, 2011

### sodaboy7

yo thanks

12. Oct 8, 2011

### Staff: Mentor

An interesting demo would be to charge up a large parallel plate capacitor with a sandwich of insulating dielectric of high permittivity, then disconnect it from the battery, and drag out the dielectric. If the dielectric's permittivity was, say 500, then the voltage on the capacitor would jump 500-fold or until the air in the gap broke down. In Tesla's early trans-Atlantic radios, I recall reading they had a couple of large barns that housed nothing but banks of large parallel plate capacitors that looked like two storeys high. The first large scale insect zappers!