# Filling a charged capacitor with dielectric material

If there is a charged capacitor (capictance $C$) connected to a battery (of EMF $V$) and the space between the plates of the capacitor is filled with a dielectric material (dielectric constant is $K$), the electric field in the region between the plates reduces by a factor of $K$.

Let $E_0$ be the electric field between the plates of the capacitor before adding the dielectric material. $$E_{\text{final}}=\frac{E_0}{K}.$$ This means $$\Delta\phi=E_{\text{final}}d=\frac{E_0}{K}d$$

($d$ is the distance between the plates of the capacitor).

But the potential difference across the capacitor should not change since it's connected to a battery of constant EMF. In this case, the potential difference reduces. Am I making any conceptual mistake?

Your initial assumption that the field is reduced by a factor $K$ is only true if the capacitor is disconnected before the dielectric is inserted. In this case, it is the charge on the plates that stays constant; the displacement field is therefore the same, but as $D= KE$, the electric field is reduced by a factor $K$.
• $E_0$ is defined to be the field before the dielectric is inserted, so it cannot "increase as you add the dielectric". Indeed, the electric field tangential to an interface is continuous, so $E_0$ would be the same as the E-field in the dielectric as it was inserted. With the voltage fixed, the E-field between the plates does not change. – Rob Jeffries May 13 '16 at 23:37