Something about parallel capacitors

AI Thread Summary
When moving parallel capacitor plates without opening the switch, the voltage remains constant while charge and electric field strength are not constant. Conversely, if the plates are moved after opening the switch, both charge and electric field strength remain constant. The work done in moving the plates affects the energy dynamics, transferring energy back to the power supply when the switch is closed. The dielectric constant remains unchanged unless the dielectric material itself is altered. This discussion highlights the differences in electrical properties based on the timing of the switch's operation relative to the movement of the plates.
illuminati23
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If two parallel plates are facing one another and one of them are moved without opening the switch, what's the difference with moving the plates after opening the switch?
 
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Sorry, You'll have to be more detailed than that.
 
If they are moved before opening, V is constant. If after, Q is constant.
 
willem2 said:
Sorry, You'll have to be more detailed than that.

So my textbook says that if the distance between the parallel plates are increased from d to d' without opening the switch first(so that means that electric current is flowing), then only V is constant and Q, E is not constant. But if d is increased to d' after opening the switch(stopping the current), then Q and E is constant.
 
Moving the plates involves doing work. That work will be transferred to potential energy or Kinetic Energy (a flow of charge), depending on whether there is a conducting path or not.
 
illuminati23 said:
So my textbook says that if the distance between the parallel plates are increased from d to d' without opening the switch first(so that means that electric current is flowing), then only V is constant and Q, E is not constant. But if d is increased to d' after opening the switch(stopping the current), then Q and E is constant.
When the switch is closed then V is the value of the power supply. So if you then increase d you put some energy back in the power supply. Increasing d means lowering the value of the capacitor. Since Q=C*V then Q will decrease. Also since E=V/d it means that E decreases. So your text reads: only V is constant, and Q and E are not constant. I think your book is a bit confusing (but correct).
The second part is OK.
 
illuminati23 said:
If two parallel plates are facing one another and one of them are moved without opening the switch, what's the difference with moving the plates after opening the switch?

-dielectric constant will decrease, capacitance will decrease, voltage between the plates will increase, and some of the charges will move back to the power supply. The energy to move the electrons back to the source will come from the work of the force that is used to depart the plates.
 
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crx said:
-dielectric constant will decrease, capacitance will decrease, voltage between the plates will increase, and some of the charges will move back to the power supply. The energy to move the electrons back to the source will come from the work of the force that is used to depart the plates.

Dielectric constant is a material property, it would not change here unless one were to change the dielectric between the plates.
 
Born2bwire said:
Dielectric constant is a material property, it would not change here unless one were to change the dielectric between the plates.

Of course it doesn't! I just wanted to see if you pay attention...:redface:
 
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