Removal of dielectric from a charged capacitor.

In summary, when we remove the dielectric from the capacitor, the extra charges flow back to the d.c. source.
  • #1
sodaboy7
81
0
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 ?
 
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  • #2
Charges will flow back through the d.c. source so that the charges stored in the capacitor will decrease to q.
 
  • #3
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
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
skmaidulhaque said:
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.

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
Delta² said:
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?

Because of the absence of the dielectric!
 
  • #7
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) Let's say we are removing the dielectric with our hands, then our hands must experience some force.
 
  • #8
sodaboy7 said:
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) Let's say we are removing the dielectric with our hands, then our hands must experience some force.

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

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
sodaboy7 said:
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??

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
yo thanks
 
  • #12
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!
 

1. How does the removal of dielectric affect the charge on a capacitor?

The removal of a dielectric from a charged capacitor will result in an increase in the electric field between the plates, which will in turn increase the potential difference and therefore the charge on the capacitor.

2. Why is it important to remove the dielectric from a charged capacitor?

Removing the dielectric from a charged capacitor is important for several reasons. It allows for a higher potential difference and therefore a higher charge to be stored on the capacitor. It also allows for better isolation and stability of the charge, as dielectric materials can sometimes break down and cause leakage currents.

3. How does the removal of dielectric affect the capacitance of a capacitor?

The capacitance of a capacitor is directly proportional to the permittivity of the dielectric material between the plates. Therefore, removing the dielectric will decrease the capacitance of the capacitor and increase its potential difference and charge.

4. Can dielectric materials be removed from a capacitor without discharging it?

No, it is not possible to remove dielectric materials from a charged capacitor without discharging it first. This is because the dielectric material is an integral part of the capacitor and its removal will drastically alter the electric field and potential difference between the plates, causing a discharge.

5. What are some common materials used as dielectrics in capacitors?

Some common materials used as dielectrics in capacitors include air, paper, ceramic, mica, and various types of plastic such as polypropylene and polystyrene. The choice of dielectric material depends on the application and desired characteristics of the capacitor.

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