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Take two capacitors that are each connected to and charged from identical constant voltage sources. The capacitors are identical in every way (distance, area, etc) except that their dielectric constants are different.

Capacitance is proportional to the dielectric constant, so all else being equal, the capacitance of the cap with higher dielectric will be greater.

According to the equation $Q = CV$, if the capacitors are both charged to (and left connected to) the same voltage then the capacitor with greater capacitance will hold more charge.

My question, assuming I have understood correctly, is why? What is physically happening on the cap with the higher dielectric that allows it to hold more charge on its plates for a given voltage? I am seeking an intuitive understanding.

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The dielectric reduces the electric field between the plates due to its ability to have induced polarization of bound charges. Therefore, you need to put more charge onto the plates than if there were no (or a weaker) dielectric in order to obtain the same potential difference between the plates. i.e. since charge will build up until the potential difference between the plates is equal to the desired voltage $V$, you get more charges with the dielectric.

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Here is an easy way to see this. Because the dielectric is polarizable, it communicates to one capacitor plate the presence of the the charge on the other more effectively than if the two plates were separated by the same thickness of a vacuum.

In this sense, you can imagine that what the dielectric does is to fool the two capacitor plates into behaving just as if they were closer together than they physically are. Since the closer together the plates are, the greater their capacitance, filling the gap between two capacitor plates with a polarizable dielectric raises their capacitance.

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  • $\begingroup$ Wouldn't the dielectric make the "communication" less effective due to the decreased field? $\endgroup$ – Aaron Stevens Nov 27 '19 at 13:05

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