Capacitor with dielectric

[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 doesn't change the electric field (V=ED), why doesn't the increase in charge on the plates increase the field?

 

[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.

 

[indigojoker]

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.

since there is an increase in charge on the plates, why doesn't 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?

 

[learningphysics]

since there is an increase in charge on the plates, why doesn't 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.

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

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

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?

yes.

 

[indigojoker]

charge increases when the dielectric is inserted so shouldn't the field increse too?

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

 

[learningphysics]

charge increases when the dielectric is inserted so shouldn't the field increse too?

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

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.

 

[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 don't 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?

 

[learningphysics]

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 don't 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?

Yes, that is correct.

 

[indigojoker]

but since the dielectric has the same amount of charge as the metal plates, wouldn't the two E-fields cancel out?

 

[learningphysics]

but since the dielectric has the same amount of charge as the metal plates, wouldn't the two E-fields cancel out?

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

 

[indigojoker]

what is the field in the dielectric?

 

[indigojoker]

would it be:

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

 

[learningphysics]

would it be:

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

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...