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Course: Physics library > Unit 12
Lesson 2: Circuits with capacitorsDielectrics in capacitors
Capacitors use non-conducting materials, or dielectrics, to store charge and increase capacitance. Dielectrics, when placed between charged capacitor plates, become polarized, reducing the voltage across the plates and increasing capacitance. The degree of capacitance increase depends on the dielectric constant of the material used. 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?(15 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.
Excellent question!
Regards,
APD(21 votes)
- Hi,
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.(7 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.(16 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)
- Hi,
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)
- If capacitance doesn't depend upon the charge and the voltage, then how does the capacitance increase when a dielectric is inserted?(2 votes)
- The dielectric changes the voltage that is required to deposit a particular charge on one of the plates. Since you now have a different relationship between Q and V than you had before, you've changed the capacitance.(3 votes)
- Does capacitor already holds charges before connecting to a battery?(2 votes)
- it could. to make sure it is 'empty', you can discharge it by touching the two plates together(3 votes)
- Can someone help me with solving questions involving dielectric slabs connected in series and parallel inside the capacitor plates. I am not able to identify if they are connected in series or parallel.(2 votes)
- If a current would have to go through both of them, they are in series.
If a current would need to choose one or the other, they are in parallel.(2 votes)
- Can you explain the concept of electric permittivity?and can the concept of relative permittivity be applied to understand dielectric ?(2 votes)
- permittivity is just a measure of how well the electric field "penetrates" through different mediums.
If we set the permittivity of free space to be 1, then we find that other materials have different permittivities and we calculate their relative permittivity by comparing to free space.(2 votes)
Video transcript
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.