If charge is constant, why does voltage increase as capacitance decrease?

Question

Consider this: we have a plate , charged with -1 C , then we put on one side a dielectric material with thickness "T", after that we put on the other side a neutral plate ( grounded) , forming a capacitor.
Now i need to calculate the voltage across the capacitor so i use C=Q/V, consider that this capacitor had a capacitance of 1 uC , so the voltage will be -1/(1*10^-6)=-1 MV.
Now we increase thickness T , decreasing capacitance , if we use C=Q/V again we will find out that as capacitance decrease , voltage increase, why?( i need a physics related explanation not math, i know they inversely proportional)

Answer 1

I'll try a simple explanation and you tell me if it makes sense.



You place a difference of 1 Coulomb of charge on two plates separated by a vacuum placed 1 meter apart. Now you force them apart to 2 meter separation. It takes work to do that because you have to apply a force to work against their mutual attractions, applied over that distance of an additional meter. That work must now be represented as a new, greater potential energy. (It's like lifting a weight against gravity, for example.) A volt is a Joule per Coulomb. The Coulombs didn't change -- still the same difference in charge. But the Joules are more, to represent the additional work you added in separating the plates. So the voltage must be higher.

Answer 2

I have answered the question i thought that you wrote but the two other respondents have interpreted it quite differently.



I interpreted that you inserted some material in the space between two plates. It couldn't be a dielectric because that would increase the capacitance.

The others have assumed that you are pulling the plates further apart.



The arguments about energy apply equally to both situations. The plates attract each other ( opposite charges attract ) To pull them apart you have to add energy.



Yet you have the same charge. A higher energy per charge is the very definition of volts.

So you have increased the volts.



And again by definition the capacitance is Q/V by definition. i.e the amount of charge that will be stored if you only have 1 V across the device.



As, by moving the plates further apart, you have more volts for the same amount of charge it must follow that if you reduce the volts to the previous level then you would have less charge ( ie lower capacitance )



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The question that I answered assumed that you had an "antidielectric"

To the best of my knowledge no such thing could exist but if it did then .............



Keeping the separation of the plates fixed and inserting the material between them,



If you can insert a material that decreases capacitance then to do so you have to force it in.

It will be repelled by the electric field.



That force and distance requires work or energy.

And the energy is distributed over the charges. The number of charges cannot alter so that the energy per charge must increase.



You can satisfy yourself of this by considering this.

To reduce the capacitance you know that you need more volts per charge.

so as the separation has not altered then the electric field must have increased.

The only way for the electric field to increase is if MORE positive charges are moved towards the positive end and more negative charges to the negative end.

And as like charges repel you would need to force this material into the space against this repulsion.



If you insert a material that increases capacitance ( a dielectric) then it gets an induced charge that causes it to be attracted by the charges on the capacitor.

It gives off energy as it enters the space ( doing work on you as you permit it to enter the gap ) so the energy per charge has reduced. ie lowered volts.



Which is completely consistent with a higher capacitance. V = Q/C

Answer 3

The plates are attracted. Increasing T requires work to be done pulling the plates apart. This increases the potential energy of the system, hence the voltage increases.



Have a look at the link.