The discussion revolves around a capacitor connected to a battery, initially holding a charge of +q and -q on its respective plates. The problem involves the insertion of a dielectric material that creates an opposing electric field of -0.30E, affecting the charge stored on the positive plate. Participants are exploring the implications of this dielectric insertion on the electric field and charge distribution.
The discussion is active, with participants offering insights into the relationships between electric fields, charge, and capacitance. Some participants are questioning assumptions about the dielectric's effect and the nature of the electric field within the capacitor, while others are attempting to clarify the equations involved.
There is uncertainty regarding whether the dielectric occupies the full gap between the capacitor plates, and participants are considering how this affects the electric field and charge calculations. The role of the battery in maintaining voltage is also a point of discussion.
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 said: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.
ehild said: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.
Why 1.7? It is wrong.Callix said: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?