If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: AP®︎/College Physics 2>Unit 3

Lesson 3: Circuits with capacitors

# Dielectrics 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.

## Want to join the conversation?

• Is there a possibility of an electric arc in the capacitor in the first part of the video?
• 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.
• Can we use a semi-conductor as dielectric?
• Hello Haroon,

Absolutely, and we do it all the time! In fact, it is the foundation of modern radio tuners.

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
• 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.
• 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.
• Does a capacitor produce AC or DC current? If so, how does it produce that type of current?
• 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.
• If capacitance doesn't depend upon the charge and the voltage, then how does the capacitance increase when a dielectric is inserted?
• 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.
• 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?
• 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??
• 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
• 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.
• Does capacitor already holds charges before connecting to a battery?
• it could. to make sure it is 'empty', you can discharge it by touching the two plates together
• 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.