Capacitors, Electric Fields, and Dielectrics

In summary, the conversation was about whether the electric field would decrease when a constant potential difference is applied to a capacitor with a dielectric inside, and how the electric field is affected by the presence of a dielectric. The conclusion was that the electric field is determined by the applied voltage and the geometry of the conductors, not by the presence of a dielectric. The formula for the electric field in a parallel plate capacitor was also mentioned, along with a website that discusses it and a potential misunderstanding by a physics teacher about the effect of a dielectric on the electric field.
  • #1
gokugreene
47
0
I have a question that is confusing me perhaps one of you can help me.
If I hook up a constant potential difference to a capacitor and place a dielectric inside of it, will the electric field decrease even if the plate separation remains constant?
I think that the capacitance will increase as well as the charge on the plates, meaning that the electric field would have to remain the same, since the potential is constant? or am I wrong? if so why?
Thanks for the help
 
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  • #2
You are correct.
 
  • #3
Do you know of anyplace that offers credible proof like in a book or an article?

This was a test question and I argued with my physics teacher about it. He thinks the electric field will decrease.. where I do not and he said I am wrong.

I would like to show him that the electric field doesn't decrease when a contstant potential difference is applied.
 
  • #4
The electric field in a parallel plate capacitor, as you apparently realize, depends only on the potential difference and the distance between the plates:
[tex]E = \frac{\Delta V}{d}[/itex]

This kind of thing should be in any textbook; here's a website that discusses it: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dielec.html#c2
 
  • #5
That is the formula I explained to him. I don't see why he doesn't get it
 
  • #6
Is it because he thought that the electric field is decreased due to the presence of the opposite electric field set up by the dielectric? But he forget that the voltage is still applied after the insertion of dielectric, which will bring the electric field back.

Pls correct me if I'm wrong.
 
  • #7
That must be what he is thinking.

DocAl, would this also apply to non parallel plate capacitors?
 
  • #8
gokugreene said:
DocAl, would this also apply to non parallel plate capacitors?
Sure. The electric field is determined by the applied voltage and the geometry of the conductors, not by the presence of a dielectric.

As Alpha2005 notes, your instructor is probably thinking of this sequence:
(1) Apply a voltage to the (empty) capacitor
(2) Remove voltage source
(3) Insert dielectric​
In this case, she would be correct. The induced polarization charge within the dielectric will reduce the effective electric field.
 

1. What is a capacitor and how does it work?

A capacitor is an electrical component that stores energy in an electric field. It consists of two conductive plates separated by a dielectric material. When a voltage is applied to the capacitor, an electric field is created between the plates, causing one plate to accumulate a positive charge and the other to accumulate a negative charge. This creates a potential difference between the plates, allowing the capacitor to store energy.

2. How do electric fields interact with conductors and insulators?

Electric fields can interact with both conductors and insulators, but in different ways. In conductors, electric fields cause the free electrons to move, creating a current. In insulators, the electric field causes the atoms to polarize, aligning in a way that resists the electric field. This is why conductors are used to carry electrical currents, while insulators are used to block or resist the flow of electricity.

3. What is the role of a dielectric material in a capacitor?

A dielectric material is an electrical insulator that is placed between the plates of a capacitor. Its role is to increase the capacitance of the capacitor by reducing the electric field between the plates. This allows the capacitor to store more charge for a given voltage. Dielectrics also help to prevent arcing between the plates, which can damage the capacitor.

4. How do you calculate the capacitance of a capacitor?

The capacitance of a capacitor is calculated using the equation C = Q/V, where C is the capacitance in farads, Q is the charge stored on the plates in coulombs, and V is the voltage across the plates in volts. The capacitance can also be calculated by multiplying the permittivity of the dielectric material by the area of the plates and dividing by the distance between the plates.

5. What are some practical applications of capacitors and electric fields?

Capacitors and electric fields have a wide range of practical applications. They are used in electronic circuits for energy storage, filtering, and timing. They are also used in power grids for power factor correction and voltage regulation. Capacitors are even used in defibrillators to deliver a high-voltage shock to the heart. Electric fields are also used in electrostatic precipitators to remove pollutants from industrial exhaust gas.

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