Something about parallel capacitors

illuminati23

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?

willem2

Sorry, You'll have to be more detailed than that.

clem

If they are moved before opening, V is constant. If after, Q is constant.

illuminati23

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.

sophiecentaur

Moving the plates involves doing work. That work will be transfered to potential energy or Kinetic Energy (a flow of charge), depending on whether there is a conducting path or not.

Per Oni

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.

crx

-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.

Born2bwire

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

crx

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