A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/2 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 C1 and without dielectric in between is C2. When the capacitor is charged by applying potential difference of V, the plate area covered by the dielectric gets charge Q1 and the rest of the area gets charge Q2. The electric field in the dielectric is E1 and that in the other portion is E2 Choose the correct option/options, ignoring edge effects.

E 1 E 2 = 1 K , V=constant , C C 2 = K + 1

A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/2 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 C1 and without dielectric in between is C2. When the capacitor is charged by applying potential difference of V, the plate area covered by the dielectric gets charge Q1 and the rest of the area gets charge Q2. The electric field in the dielectric is E1 and that in the other portion is E2 Choose the correct option/options, ignoring edge effects.

C 1 = K ∈ 0 A 2 d C 2 = ∈ 0 A 2 d C = C 1 + C 2 = K ∈ 0 A 2 d + ∈ 0 A 2 d = ∈ 0 A 2 d [ K + 1 ] C = C 2 ( K + 1 ) C C 2 = K + 1 V is same for both C 1 and C 2 E 1 E 2 = 1

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A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/2 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 C₁ and without dielectric in between is C₂. When the capacitor is charged by applying potential difference of V, the plate area covered by the dielectric gets charge Q₁ and the rest of the area gets charge Q₂. The electric field in the dielectric is E₁ and that in the other portion is E₂ Choose the correct option/options, ignoring edge effects.

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

V=constant

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

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

answer is A, B, C.

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Detailed Solution

$C_{1} = \frac{K \in_{0} A}{2 d} C_{2} = \frac{\in_{0} A}{2 d}$
$\begin{array}{l} C = C_{1} + C_{2} = \frac{K \in_{0} A}{2 d} + \frac{\in_{0} A}{2 d} = \frac{\in_{0} A}{2 d} [K + 1] \\ C = C_{2} (K + 1) \\ \frac{C}{C_{2}} = K + 1 \end{array}$
V is same for both $C_{1} and C_{2}$
$\frac{E_{1}}{E_{2}} = 1$