# Capacitor with dielectric

1. Oct 11, 2007

### indigojoker

there is a capacitor with a constant potential difference of V in air. When a dielectric is completely inserted between the plates of the capacitor completely filling it, additional charge flows onto the positive plate.

The additional charge should be Q=kCV-CV right?

Would the induced charge on either faces of the dielectric be Q=kCV?

Since the dielectric doesnt change the electric field (V=ED), why doesnt the increase in charge on the plates increase the field?

Last edited: Oct 11, 2007
2. Oct 11, 2007

### learningphysics

Yes, kCV is correct. extra charge is kCV - CV.

V = Ed

V is the same. d is the same. So E must be the same.

3. Oct 11, 2007

### indigojoker

since there is an increase in charge on the plates, why doesnt this give rise to an increase in the E field since the increase in the chrage on the plates should serve to increase the field between the plates? I know V is the same and so E must be the same, I'm looking more for a conceptual idea.

also, there is kCV charge on the top on the dielectric right? does that mean that there is a -kCV charge on the bottom of the dielectric?

4. Oct 11, 2007

### learningphysics

Well, the field is $$\frac{\sigma}{\epsilon}$$ which is $$\frac{Q}{A\epsilon}$$

so how does this change when the dielectric is inserted...

yes.

5. Oct 11, 2007

### indigojoker

charge increases when the dielectric is inserted so shouldnt the field increse too?

to: $$\frac{kCV}{A\epsilon}$$

6. Oct 11, 2007

### learningphysics

k is the dielectric constant. so $$\epsilon = k\epsilon_0$$

plug this in and the k's cancel in the numerator and denominator...

so you're left with

$$\frac{CV}{A\epsilon_0}$$, which is just like without the dielectric.

7. Oct 11, 2007

### indigojoker

okay, so i want to understand this in a non mathematical way.

in the same situation as above, we insert a dielectric and then we get an increase in charge on the plates. the reason why we dont get an increase in the E-field is because the dielectric creates an opposing field that makes it so the E-field does not change. is the right?

8. Oct 11, 2007

### learningphysics

Yes, that is correct.

9. Oct 11, 2007

### indigojoker

but since the dielectric has the same amount of charge as the metal plates, wouldnt the two E-fields cancel out?

10. Oct 11, 2007

### learningphysics

No, just because the charge is the same doesn't mean the field is the same.

11. Oct 11, 2007

### indigojoker

what is the field in the dielectric?

12. Oct 11, 2007

### indigojoker

would it be:

$$E=\frac{kQ}{A\epsilon_o}-\frac{Q}{A\epsilon_o}$$

13. Oct 11, 2007

### learningphysics

Yeah... I would write it as: $$E=\frac{Q}{A\epsilon_o}-\frac{kQ}{A\epsilon_o}$$

so when you add the field due to just the charge... you get the net field...