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(C) The capacitor is connected to a battery of constant potential difference $V$. Since the plates are connected directly to the battery,the potential

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electric fields at points $P$ and $P'$ are equal.

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(C) The capacitor is connected to a battery of constant potential difference $V$. Since the plates are connected directly to the battery,the potential difference across any vertical cross-section of the capacitor must be equal to the potential difference of the battery,which is $V$. For a parallel plate capacitor,the electric field $E$ between the plates is related to the potential difference $V$ and the separation $D$ by the relation $V = E \cdot D$,or $E = \frac{V}{D}$. This relation holds true for both the region filled with the dielectric (point $P$) and the region filled with air (point $P'$),because the potential difference across the plates is uniform throughout the capacitor. Therefore,the electric field at both points $P$ and $P'$ is equal to $\frac{V}{D}$.

$A$ parallel plate capacitor,partially filled with a dielectric slab of dielectric constant $K$,is connected to a cell of emf $V \text{ volt}$,as shown in the figure. The separation between the plates is $D$. Then:

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