Two 5.0 mm x 5.0 mm electrodes with a 0.10 mm-thick sheet of Mylar between them are attached to a 9.0 V battery. Without disconnecting the battery, the Mylar is withdrawn. (Very small spacers keep the electrode separation unchanged.) Dielectric constant for Mylar is kMylar=3.1.
What is the charge before and after the Mylar is withdrawn?
What is the potential difference before and after the Mylar is withdrawn?
What is the electric field before and after the Mylar is withdrawn?

Question

Two 5.0 mm x 5.0 mm electrodes with a 0.10 mm-thick sheet of Mylar between them are attached to a 9.0 V battery. Without disconnecting the battery, the Mylar is withdrawn. (Very small spacers keep the electrode separation unchanged.) Dielectric constant for Mylar is kMylar=3.1.
What is the charge before and after the Mylar is withdrawn? 
What is the potential difference before and after the Mylar is withdrawn? 
What is the electric field before and after the Mylar is withdrawn?

anonymous · Answer

I know you first us $$C=\epsilon_0A/d$$ Then you us Q=CV. But how do you include the kMylar?

anonymous · Answer

us is suppose to say use

kainui · Answer

Since you're not disconnecting the battery you will have a constant source of charge depending on if you gain or lose capacitance. What do dielectrics do? First off you need to understand what a capacitor does. The capacitor stores positive charge on one side and negative charge on the other side depending on how much voltage is applied. That's why the unit for capacitance, the farad, is the same thing as coulombs per volt. So the more volts of potential you have, the more charge you can store. However, there's a limit as to how much charge a capacitor can hold before too many volts of potential give it too much of a charge that the electrons jump through the air and neutralize the charge. That's where dielectrics come in to prevent that and thereby allow the capacitor to hold more charge.

So that explains more or less why a dielectric increases the capacitance of a capacitor, but now how can we relate that to a formula? Simply, the dielectric increases the capacitance proportionally based on the dielectric constant. C(with dielectric) = kC.

What is the potential difference before and after? Since it's never disconnected from the battery, it should stay constant. It will only change if you remove the plate while it's disconnected from a battery.

Remember that voltage is just electric field times distance.

If you need more help I'd be happy to try to steer you in the right direction. =)