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New Capacitance in Dielectric

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Question

Figure (i) shows a capacitors of capacitance $C_{1}$ . A dielectric of dielectric constant K filling half the volume between the plates are inserted as shown in (ii) and (iii) having capacitances $C_{2}$ and $C_{3}$ . Then,

C_{1} = C_{2} = C_{3}

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C_{2} < C_{1} a n d C_{3} > C_{1}

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C_{2} > C_{1} a n d C_{3} > C_{1}

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C_{2} > C_{1} a n d C_{3} < C_{1}

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Solution

The correct option is D $C_{2} > C_{1} a n d C_{3} > C_{1}$
For (i), $C_{1} = \frac{A ϵ_{0}}{d}$
For (ii), here two capacitors are in series- one is air filled capacitor with separation $d / 2$ and other is dielectric filled with separation $d / 2$ . Thus capacitance one is $2 C_{1}$ and other is $2 k C_{1}$ where k is dielectric constant.
$C_{2} = \frac{2 C_{1} .2 k C_{1}}{2 C_{1} + 2 k C_{1}} = \frac{2 k C_{1}}{k + 1}$
For (iii), here two capacitors are in parallel- one is one is air filled capacitor with area $A / 2$ and other is dielectric filled with area $A / 2$ .Thus capacitance one is $C_{1} / 2$ and other is $k C_{1} / 2$
$C_{3} = C_{1} / 2 + k C_{1} / 2 = \frac{k + 1}{2} C_{1}$
As $k > 1, C_{2} > C_{1}$ and $C_{3} > C_{1}$

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Figure (i) shows a capacitors of capacitance C1. A dielectric of dielectric constant K filling half the volume between the plates are inserted as shown in (ii) and (iii) having capacitances C2 and C3. Then,

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