A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/3 of the area of its plates, as shown in the figure. The total capacitance of the capacitor is C while that of the portion with dielectric in between is $C_{1} .$ When the capacitor is charged, the plate area covered by the dielectric gets charge $Q_{1}$ and the rest of the area gets charge $Q_{2}$ .The electric field in the dielectric is $E_{1}$ and that in the other portion is $E_{2}$ . Choose the correct option/options, ignoring edge effects

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$\frac{E_{1}}{E_{2}} = \frac{1}{K}$

$\frac{C}{C_{1}} = \frac{2 + K}{K}$

$\frac{Q_{1}}{Q_{2}} = \frac{3}{K}$

$\frac{E_{1}}{E_{2}} = 1$

answer is A, D.

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Detailed Solution

$∵ ΔV$ across dielectric as wall as air regions is same, $E_{1} = E_{2}$
$C_{1} = \frac{K \in_{0} A / 3}{α} = \frac{K \in_{0} A}{3 d} = K (\frac{\in_{0} A}{3 d}) C = \frac{K \in_{0} A}{3 d} + \frac{2 \in_{0} A}{3 α} = K (\frac{K + 2}{3}) \frac{\in_{0} A}{2} \Rightarrow \frac{C}{C_{1}} = \frac{K + 2}{K} Q_{1} = C_{1} Δ C V; Q_{2} = \frac{2}{3} (\frac{\in_{0} A}{d}) Δ V \frac{Q_{1}}{Q_{2}} = \frac{(\frac{K \in_{0} A}{3 d})}{\frac{2 \in_{0} A}{3 α}} = \frac{K}{2}$
$\Rightarrow A & D$ are correct.

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