A parallel plate capacitor has a dielectric slab of dielectric...

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.

A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/3 of the

Options

A $\dfrac{{E}_{1}}{{E}_{2}} = 1$
B $\dfrac{{E}_{1}}{{E}_{2}} = \dfrac{1}{K}$
C $\dfrac{{Q}_{1}}{{Q}_{2}} = \dfrac{3}{K}$
D $\dfrac{C}{{C}_{1}} = \dfrac{2 + K}{K}$

Correct answer

A. $\dfrac{{E}_{1}}{{E}_{2}} = 1$

Step-by-step solution

E = \dfrac{v}{d}

{E}_{1} = \dfrac{v}{d}

{E}_{2} = \dfrac{v}{d}

(As both parts have some P.D.) upper capacitor have capacitance

{C}_{1} = K{&epsilon;}_{0}A / d

Lower capacitor

{C}_{2} = 2{&epsilon;}_{0}A / d

&there4;

equivalent capacitance

C = \left(k + 2\right){ϵ}_{0}A / d

&there4; \dfrac{C}{{C}_{1}} = \dfrac{k + 2}{k}

And

{E}_{1}: {E}_{2} = \dfrac{v}{d}: \dfrac{v}{d} = 1

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