Understanding the capacitor deeply

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SUMMARY

This discussion focuses on the behavior of capacitors, specifically addressing the equal charge distribution on plates A and B when a voltage is applied, and the relationship between frequency and current flow. The Conservation of Charge Law explains why the charges on both plates are equal in magnitude. Additionally, it is established that at higher frequencies, the impedance of the capacitor decreases, allowing for greater current flow, as the current through a capacitor is determined by capacitance and the rate of voltage change.

PREREQUISITES
  • Understanding of basic capacitor operation and structure
  • Familiarity with the Conservation of Charge Law
  • Knowledge of AC voltage behavior in capacitors
  • Concept of impedance in electrical circuits
NEXT STEPS
  • Study the relationship between capacitance and frequency in capacitors
  • Learn about the mathematical representation of capacitive reactance (X = -j/wC)
  • Explore the effects of dielectric materials on capacitor performance
  • Investigate the implications of charge density changes in AC circuits
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Electrical engineers, physics students, and anyone interested in deepening their understanding of capacitor behavior in AC circuits.

finenil
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hello every body,
i am trying to understand the capacitor behavior as deeply as i can ...
consider a simple capacitor with two plates A and B

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1- why the charge on both plates are equal in magnitude ?
when we apply a voltage to a capacitor,charge will accumulate on side A.this charge will build an E field which will pass thru the insulator and exert force on the plate B.So charges on plate B move and accumulate together.Ok but how do we know that this E field caused by Charges on plate A will exactly attract the same amount of charge that was on plate A on plate B??
why these two has to be equal ? Conservation of Charge Law ?

2 - why is that the higher the frequency , the more is the current ? is the following intuition wrong ?
when we apply a ac voltage to plate A of a capacitor , charges density on plate A will change with voltage.
when charge density changes , E field caused by them in the Capacitor changes too.
This changing E filed will cause the plate B charge density changing .
Ok but I am thinking that when the E filed is changing very fast(high freq.) the plate B charges may not find enough time to respond to it and move to the new configuration.so the charge density(voltage) on plate B may lag the charge density(voltage) of plate A. So why is that at high freq. the capacitor acts like a short curcuit or the impedance becomes lower?:confused:
 
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simple answer:afaik the current through a capacitor depends only on the capacitance and the rate of change of voltage. (i assume that this is because the total charge stored in the capacitor depends only on the voltage).

more complex answer:i think the capacitance may depend on the frequency though. i don't know anything about that.
 
granpa said:
more complex answer:i think the capacitance may depend on the frequency though. i don't know anything about that.

capacitance depends only on geometrical aspects + the electrical field permitivity of the insulator between the plates.The insulator characteristics may generally change with frequency but that is not our discussion.
 
1.When voltage is applied to the capacitor, positive and negative charges go onto different plates. Say positive charge on plate A and negative one on plate B.

Allright, let's imagine you somehow inject electrons into plate A of the capacitor, these electrons will repel the electrons hidden within the other uncharged plate, right? And these other electrons will be forced out of the B plate of the capacitator. However, if the other capacitor plate isn't connected to anything, then those electrons cannot leave the other plate! They will continue to repel the electrons in plate A, thus very few electrons can be forced into plate A. This means that if charge is injected into capacitor's plate A, then an equal amount of charge is pushed out of the capacitor's plate B.

Anyway, that's kind of the way I understand this, although a little childish it might as well be correct :)

2. " Ok but I am thinking that when the E filed is changing very fast(high freq.) the plate B charges may not find enough time to respond to it and move to the new configuration.so the charge density(voltage) on plate B may lag the charge density(voltage) of plate A. So why is that at high freq. the capacitor acts like a short curcuit or the impedance becomes lower? "
I agree with you. The higher the frequence the smaller is the resistance of the capasitor X=-j/wC (j - complex number) and the explanation you give seems reasonable, I can't really think of anything better.

cheers
 

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