How Does Inserting a Dielectric Affect Electric Field in a Charged Capacitor?

[kliang1234]

Information given:
A parallel-plate capacitor has capacitance C = 12.5pF when the volume between the plates is filled with air. The plates are circular, with radius 3.00cm. The capacitor is connected to a battery and a charge of magnitude 25.0pC goes onto each plate. With the capacitor still connected to the battery, a slab of dielectric is inserted between the plates, completely filling the space between the plates. After the dielectric has been inserted the charge on each plate has magnitude 45.0pC.


Question:
1) What is the electric field at a point midway between the plates before the dielectric has been inserted?

2)What is the electric field at a point midway between the plates after the dielectric has been inserted?



Apparently, the correct answer for both questions is 999V/m.
The answer for the first part is calculated with E=Q/Ae

The justification for #2 is that because the plates are still connected to the battery, potential does not change, therefore by E=V/d, the electric field does not change after the dielectric is inserted.


This is where I'm confused. I understand their justification that if potential doesn't change, electric field doesn't change.
However, the given information specifically states that the charge increases from 25pC to 45pC. From what i know, such an increase in charge should increases electric field.

Lets assume that i attempted problem 2 with the same approach i took problem 1.
E=Q/Ae. The increase in Q increases electric field if that equation was used.

Can anyone explain this to me. I'm very confused.

Thanks

 

[eczeno]

i think i can clarify a bit.

you have it right that for electrostatic fields, E=V/d, so since the potential has not changed the (total) E-field must be the same as before.

So, how do we get more charge with the dielectric?

well, the dielectric is an insulator, so it does not have free charges; however, when placed in an electric field, the bound charges in the atoms separate slightly, creating their own electric field, which opposes the external one.

thus the total electric field, if we measured it at the center, would be $E_{capacitor} - E_{dielectric}$. This is what must equal V/d, which means $E_{capacitor}$ must be bigger than before, and thus more charge is required.

Lets look at a similar situation which might help in understanding. Imagine we charge the capacitor, but then disconnect it from the battery before putting the dielectric in. now, the charge on the plates cannot change, but the potential difference can change. what will happen?

well, since the charge is fixed, and the dielectric sets up an opposite field, the total field inside the capacitor must go down. and indeed it does. this time however, the charge cannot change to pump the E-field back up, so the potential must also go down for E=V/d to hold. and indeed it does too.