A parallel plate capacitor has a dielectric slab of dielectric constant K between...

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.

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

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

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

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

Solution:

E = \frac{v}{d}

E_{1} = \frac{v}{d}

E_{2} = \frac{v}{d}

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

C_{1} = K ε_{0} A / d

Lower capacitor

C_{2} = 2 ε_{0} A / d

∴

equivalent capacitance

C = (k + 2) ϵ_{0} A / d

∴ \frac{C}{C_{1}} = \frac{k + 2}{k}

And

E_{1} : E_{2} = \frac{v}{d} : \frac{v}{d} = 1

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A parallel plate capacitor has a dielectric slab of dielectric constant K between...

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IIT Jodhpur

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A parallel plate capacitor has a dielectric slab of dielectric constant K between...

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