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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?

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With the capacitor remaining connected to the battery, then the potential difference between the plates is unchanged. If the potential difference stays the same and the gap between the plates stays the same, then the electric field stays the same.

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 $K$.

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The first equation assumes the external electric field (caused by the charges on the plates of the capacitor) doesn't change. When a battery is connected, it can fill and discharge the plates as necessary to maintain the voltage. So, the E_0 value increases as you add the dielectric.

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  • $\begingroup$ $E_0$ 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. $\endgroup$ – Rob Jeffries May 13 '16 at 23:37

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