Capacitor Charge Calculation with Dielectric Insertion

[ehild]

Well logically I would suppose not then, since C=q/V and we already know V is constant because of the battery, so q must change. Adding dielectrics help increase C, so that means q increases as well.

So before, were you saying that E=E1-(-E2)? Or E=E1+(-E2), because it seems like you're hinting that it's the first, which in that case it would be by a factor of 1.7.

It is the first, E1=E+E2, but E2 is the field of the dipoles, which is 0.3E, if the E2 means the magnitude of the dipole field.

 

[Callix]

It is the first, E1=E+E2, but E2 is the field of the dipoles, which is 0.3E, if the E2 means the magnitude of the dipole field.

Right, so then it would increase by a factor of 1.7, so
σ'=1.7σ=1.7ε₀E₁.
And that is the surface charge density on the dielectric?

 

[ehild]

Right, so then it would increase by a factor of 1.7, so
σ'=1.7σ=1.7ε₀E₁.
And that is the surface charge density on the dielectric?

Why 1.7? It is wrong.