physical explanation question.. - Electronic Circuits Projects Diagrams Free

transgalactic · 3rd November 2009, 07:41 PM

why when when we connect capacitators in a row(not paralel)
the capacitance gets smaller

i need the physical explanation for it

birdman0_o · 3rd November 2009, 08:27 PM

Quote:

The figure to the right shows two capacitor symbols connected in series. As a starting point, let's assume that these are two identical capacitors. The connection between them is assumed to have no resistance, and therefore no effect on the behavior of these to capacitors or any circuit in which they may be connected. Therefore, this connection may be any length, covering any distance, without having any noticeable effect.
Series capacitors with no separation between them.

This being the case, let's shorten the distance between capacitors to zero. This means that the connected plates of the two capacitors will actually touch, as shown in the second image to the right.

Next, we recognize that the thickness of that center plate is unimportant; it's simply a broad conductor between the two capacitors. Therefore we can make this center plate as thin as we want. Therefore, at least in theory, we can reduce it to atomic thickness without any effect on the capacitance of the series combination.
Series capacitors with the center plate removed.

But that center plate is nothing more than an equipotential plane in the middle of an electric field. Since the outer plates are still parallel to each other, removing the center plate won't change the total electric field. This leaves us with a single capacitor, but with the plates spaced twice as far apart as for either of the original capacitors. As a result of this, the combined capacitance of the two identical capacitors in series is just half the capacitance of either one.

Capacitors in Series

rajbex · 3rd November 2009, 09:02 PM

Lets take an example of two equal value parallel plate capacitors. The capacitance is directly proportional to the amount of charge it is holding and inversely proportional to the voltage applied that is pushing the charge onto the opposite plates. Remember, the net amount of charge a capacitor holds is always zero, but we talk about the numerical value of +ive (or the equal -ve) charge as the charge stored by the capacitor. When we connect capacitors in series, we are connecting a +ive terminal of one to the -ive terminal of other that results in the summing up the potential differences, and the net voltage across the connection is the sum of individual voltages. However, when connecting a +ive to a -ive terminal, the opposite charges get cancelled, and you will be left with the same amount of charge as in a single capacitor. So the situation is like you require more potential difference to hold the same amount of charge, which means less capacitance (C=Q/V).

- Raj
Experiments with PIC16F628A

transgalactic · 3rd November 2009, 10:17 PM

i agree with this line
"Therefore, at least in theory, we can reduce it to atomic thickness without any effect on the capacitance of the series combination"

but i cant understand why we can remove the center plate??
"Since the outer plates are still parallel to each other, removing the center plate won't change the total electric field"

why the iron plate is not changing the total electric field?

i know that when we put dialectric plate in a capacitor the capacitance changes

birdman0_o · 4th November 2009, 05:02 AM

Having a positive and negative charge next to each other virtually cancels them out. Capacitance is based on charge separation over a distance. We do not make up the laws of physics, we follow them

transgalactic · 4th November 2009, 06:23 AM

ahh the positive and negatie charges cancel each other in the added plates
so the is no chrge and no electric field of this center plate.

and we have one capacitor with twice the distance
of each and thecapacitance gets smaller by half too

thanks

birdman0_o · 4th November 2009, 05:25 PM

No problem, just went over this in class today actually (Review from High School)

transgalactic · 11th November 2009, 07:45 PM

my prof said that its not working for non lenear capacitators
(it works only for a tiny central plate)

he said that the correct answer is:
the total charge is constant
and
because q=cu

i dont understand this logic

your explanation was much clearer
i cant see how the total capacitance reduces

birdman0_o · 11th November 2009, 08:21 PM

If the potential is fixed he is wrong.

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3rd November 2009, 07:41 PM	#1
transgalactic	physical explanation question.. why when when we connect capacitators in a row(not paralel) the capacitance gets smaller i need the physical explanation for it

3rd November 2009, 09:02 PM	#3
rajbex	Lets take an example of two equal value parallel plate capacitors. The capacitance is directly proportional to the amount of charge it is holding and inversely proportional to the voltage applied that is pushing the charge onto the opposite plates. Remember, the net amount of charge a capacitor holds is always zero, but we talk about the numerical value of +ive (or the equal -ve) charge as the charge stored by the capacitor. When we connect capacitors in series, we are connecting a +ive terminal of one to the -ive terminal of other that results in the summing up the potential differences, and the net voltage across the connection is the sum of individual voltages. However, when connecting a +ive to a -ive terminal, the opposite charges get cancelled, and you will be left with the same amount of charge as in a single capacitor. So the situation is like you require more potential difference to hold the same amount of charge, which means less capacitance (C=Q/V). - Raj Experiments with PIC16F628A

3rd November 2009, 10:17 PM	#4
transgalactic	i agree with this line "Therefore, at least in theory, we can reduce it to atomic thickness without any effect on the capacitance of the series combination" but i cant understand why we can remove the center plate?? "Since the outer plates are still parallel to each other, removing the center plate won't change the total electric field" why the iron plate is not changing the total electric field? i know that when we put dialectric plate in a capacitor the capacitance changes

4th November 2009, 05:02 AM	#5
birdman0_o	Having a positive and negative charge next to each other virtually cancels them out. Capacitance is based on charge separation over a distance. We do not make up the laws of physics, we follow them __________________ Mike My website: www.ElectroBird.net

4th November 2009, 06:23 AM	#6
transgalactic	ahh the positive and negatie charges cancel each other in the added plates so the is no chrge and no electric field of this center plate. and we have one capacitor with twice the distance of each and thecapacitance gets smaller by half too thanks

4th November 2009, 05:25 PM	#7
birdman0_o	No problem, just went over this in class today actually (Review from High School) __________________ Mike My website: www.ElectroBird.net

11th November 2009, 07:45 PM	#8
transgalactic	my prof said that its not working for non lenear capacitators (it works only for a tiny central plate) he said that the correct answer is: the total charge is constant and because q=cu i dont understand this logic your explanation was much clearer i cant see how the total capacitance reduces Last edited by transgalactic; 11th November 2009 at 07:45 PM.

11th November 2009, 08:21 PM	#9
birdman0_o	If the potential is fixed he is wrong. __________________ Mike My website: www.ElectroBird.net