How dielectrics function in circuits. By David Santo Pietro. . Created by David SantoPietro.
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- Is there a possibility of an electric arc in the capacitor in the first part of the video?(22 votes)
- If the voltage applied across any capacitor becomes too great, the dielectric will break down (known as electrical breakdown) and arcing will occur between the capacitor plates resulting in a short-circuit. The working voltage of the capacitor depends on the type of dielectric material being used and its thickness.(29 votes)
- Can we use a semi-conductor as dielectric?(14 votes)
- Hello Haroon,
Absolutely, and we do it all the time! In fact, it is the foundation of modern radio tuners.
Please see https://en.wikipedia.org/wiki/Varicap
Here a diode (semiconductor PN junction) operates as a capacitor with the ability to change capacitance based on the applied voltage. In radio terms this allows us to make voltage controlled oscillators.
I have a question and would like to seek help in clarifying it.
Let's say I have 2 different scenarios:
Scenario 1: a capacitor, without a dielectric material, was charged with a battery and the battery was removed after charging the capacitor fully.
Scenario 2: a capacitor, with a dielectric material, was charged fully with a battery that has the same voltage in scenario 1. The battery was also taken away after charging fully.
In scenario 2, the capacitance would have increased due to the increase in charge accumulated on the plates (as explained in the video). However, if we were to compare the voltage across the plates, between scenario 1 & 2, does this imply that the voltage across the plates in scenario 2 is smaller than that in scenario 1, when the batteries were removed?
Please kindly help to clarify this doubt. Thank you.(6 votes)
- I'm sorry but JV Fernandez's answer is incorrect. The voltage across the capacitors in both the scenarios would be the same. This makes sense because they were both charged to the same voltage initially. However the capacitor with the dielectric holds more charge than the one without a dielectric.
I also disagree with David on what happens when we insert a dielectric after the capacitor is charged and unplugged from the battery. Keep in mind that always C=Q/V. He says the voltage goes down, which causes the capacitance to increase. It is not so, but actually vice versa. This comes from the fact that C=ε*A/d where ε is the permittivity of the dielectric. The capacitance increases but the charge has nowhere to go, thus causing the voltage to decrease. Hope this helps.(14 votes)
- Does a capacitor produce AC or DC current? If so, how does it produce that type of current?(3 votes)
- If there are no other external power supplies, a discharging capacitor will produce a DC current, similar to that of a battery except the voltage decreases as the capacitor discharges.(6 votes)
- Why didn't the remaining 3 positive charges on the left side of the capacitor stay instead of moving towards the negative 3 charges on the right side of the capacitor when connected to the voltage (battery)? Furthermore, why does the battery have to increase the number of charges on the capacitors in order to make equivalent the voltage?(2 votes)
- good questions
the positive charges do not move. this is because the positive charge is carried by the protons. The protons are FIXED in position, in the nucleus of the atom. Only the electrons can move in this circuit and they are negatively charged. This is why only negative charges moved.
the battery will 'pump' the electrons around the circuit trying to increase the number of electrons on the negative plate. The electrons feel a force inviting them to return to the positive plate (back through the battery) but the battery keeps pushing against that force. When the voltage (push) between the plates is as big as the push from the battery, then the battery can not move any more electrons... so this is why the final voltage across the capacitor (when it has finished charging up) is the same as the voltage across the battery.
this make sense ok??(6 votes)
I may be wrong but you argued that a dielectric would help increase the capacitance by decreasing the voltage. I think that is wrong. Capacitance is a constant value which stays the same. If we increase the charge the voltage would also increase to keep the capacitance constant.
So, when we introduce the dielectric or any insulating material, it experiences a partial ionization that permits conduction through it. This is called dielectric breakdown. Many dielectric materials can tolerate stronger electric fields without breaking down. Thus using dielectric allows a capacitor to sustain a higher potential difference and so greater amount of charge and energy.
Could you please clarify if I am wrong or if my concept is based on different assumptions.
Another argument is that V= line integral (E. dL). So if we somplify the form we will get V= q/(4*pi*epsilon*r). The ratio proportion b/t V and q is a direct proportion. Therefore, by keeping the charge same we cannot increase the electric potential.
My source of information is from "University Physics" 12th edition by Hugh D. Young and Roger A. Freedman. And the second paragraph in my comment is paraphrased by me to explain the concept. It was taken from pagfge 828 topic number 24.4(2 votes)
- Capacitance is a constant value of the capacitor but that constant is determined by the dielectric (among other things). Parallel plates with a vacuum between them will have less capacitance than the same plates with a dielectric between them.(3 votes)
- Why do you have to do work to insert a dielectric into a capacitor that's not connected to a battery?(2 votes)
- The potential energy stored in a capacitor is PE = 1/2CV² = 1/2Q²/C.
For the case of a capacitor not connected to a battery, the charge Q will be fixed.
Adding a dielectric will increase the capacitance C and therefore decrease the potential energy stored in the capacitor, so you have to do negative work to decrease that potential energy.
What's interesting about this problem is if the voltage if fixed rather than the charge (ie. battery is connected), then you would need to do positive work to insert the dielectric.(2 votes)
- Does physical conditions(like temperature, pressure etc.) play role in affecting both dielectrics and Capacitance ?(1 vote)
- Capacitance is only a function of the dielectric constant of the insulator and the physical dimensions of the capacitor, at least what is modeled in circuit simulators like Spice. However, experiments have shown that polarization can be affected by temperature and pressure which will have an effect on the dielectric constant of the insulator, but these effects are ignored in practical applications.(3 votes)
- i have seen some light on capacitor but hopeless in transistor. Can anybody help to explain the working principle of npn and pnp transistor(2 votes)
- Hello Szvon,
In both cases a small current in the "base" of the transistor controls a larger current that flows from "emitter" to "collector." The transistor acts as an amplifier.
Recommend searching the web for videos about transistors. While you are at it research op-amps. When you start building circuits op-amps are much easier to use...
Welcome to electronics!
- When a dielectric slab. Is gradually inserted between the plates of an isolated parallel plate capacitor , the energy of the system decreses. What can you conclude about the force on the slab exerted by the electric field ?
Please help soon.
Thank you.(1 vote)
- As the dielectric slab is being inserted, the space inside the capacitor filled with the dielectric is increasing with time, which means that the electric field must perform positive work on the dielectric slab, making up a force that pulls the piece inside. You must remember that force is the rate of change of energy over time, calculated by: F = -dU/dt. If the energy decreases with time, the strength also decreases.(3 votes)
In most capacitors, a non-conducting material is placed between the two metal pieces that make up that capacitor. There's two reasons for this. For one, the non-conducting material prevents the pieces of metal from touching each other, which is important because if the pieces of metal were touching, no charge would ever get stored since you've completed the circuit. But there's another bonus to inserting a non-conducting material between the plates of a capacitor. It will always increase the capacitance of that capacitor. As long as the material is non-conducting, it doesn't even matter what it is. As long as you don't change the area or separation between the plates, inserting a non-conducting material will always increase the capacitance. The name we give to non-conducting materials place between capacitor plates is a dielectric. But why does a dielectric increase the capacitance? To find out, let's look at this example. When you hook up a battery of voltage V to a capacitor, charge will get separated. Now let's say you remove the battery. The charge is stuck on the plate since the negatives don't have a path in which to get back to the positives. So even after removing the battery, the charge on the plates is going to remain the same. And the voltage will also remain the same as the voltage of the battery that charged it up. Now imagine placing a dielectric in between the plates of the capacitor. The dielectric material is made out of atoms and molecules, and when placed in between the plates of this charged up capacitor, the negative charges in the dielectric are going to get attracted to the positive plate of the capacitor. But those negatives can't travel to the positive plate since this dielectric is a non-conducting material. However, the negatives can shift or lean towards the positive plate. This causes the charge in the atoms and molecules within the dielectric to become polarized. To put it another way, the atom kind of stretches and one end becomes overall negative and the other end becomes overall positive. It's also possible that the dielectric material started off polarized because some molecules are just naturally polarized like water. In this case, when the dielectric is placed between the charged up capacitor plates, the attraction between the negative side of the polarized molecule and the positive plate of the capacitor would cause the polarized molecules to rotate, allowing the negatives to be a little bit closer to the positively charged capacitor plate. Either way, the end result is that the negatives in the atoms and molecules are going to face the positive capacitor plate and the positives in the atoms and molecules are going to face the negative capacitor plate. So how does this increase the capacitance? The reason this increases the capacitance is because it reduces the voltage between the capacitor plates. It reduces the voltage because even though there's still just as many charges on the capacitor plates, their contribution to the voltage across the plates is being partially cancelled. In other words, some of the positive charges on the capacitor plate are having their contribution to the voltage negated by the fact that there's a negative charge right next to them now. Similarly, on the negative side there's just as much negative charge as there ever was, but some of the negative charges are having their contribution to the voltage canceled by the fact that there's a positive charge right next to them. So the total charge on this capacitor has remained the same, but the voltage across the plates has been decreased because of the polarization of the dielectric. If we look at the definition of capacitance, we see that if the charge stays the same and the voltage decreases, the capacitance is going to increase, because dividing by a smaller number for the voltage is going to result in a larger value for the capacitance. So inserting a dielectric in this case, increase the capacitance by lowering the voltage. Let's look at another case of inserting a dielectric. Imagine we, again, let a battery of voltage V fully charge this capacitor. And let's insert a dielectric between the plates. But this time, let's leave the battery connected. Now what's going to happen? Well, just like before, the atoms and molecules in the dielectric are going to stretch and orient themselves so that the negatives are facing the positive plate and the positives are facing the negative plate, which again reduces the voltage between the two capacitor plates. But remember, we left the battery connected and this battery is going to try to do whatever it has to do in order to make sure the voltage across the capacitor is the same as the voltage of the battery V. Because that's just what batteries do. They try to maintain a constant voltage. So since the dielectric reduced the voltage by canceling the contributions from some of the charges, the battery's just going to cause even more charges to get separated until the voltage across the capacitor is again the same as the voltage of the battery. So the charge stored on the capacitor is going to increase, but the voltage is going to stay the same. Looking at the definition of capacitance, the charge on the capacitor increased after we inserted the dielectric. But the voltage across the capacitor plates stayed the same, since it's still hooked up to the same battery. So the effect of inserting a dielectric again is to increase the capacitance, this time by storing more charge for the same amount of voltage. To figure out how much you've increased the capacitance, you just need to know what's called the dielectric constant of the material that you've inserted between the capacitor plates. The dielectric constant is often represented with a Greek letter kappa or simply a K. The formula for finding out how the dielectric will change the capacitance is simple. If the capacitance of a capacitor before inserting a dielectric was C, then the capacitance after inserting a dielectric is just going to be k times C. We should note that since a dielectric always increases the capacitance, the dielectric constant k for a non-conducting material is always greater than 1. So for example, if a capacitor as a capacitance of 4 farads, when you insert a dialect with dielectric constant 3, the capacitance will become 12 farads.