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Polarization Vector

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Question

Two conducting parallel plates have area $A$ , are separated by distance dd, and are filled with a dielectric material of constant $_{1}$ .The capacitance of this system is $C$ . A student plans to replace the dielectric $_{1}$ with a second dielectric, $_{2}$ , where $_{2} = 2_{1}$ .However, the student only has half the volume of dielectic material to fully replace $_{1}$ . So the student fills half the gap with $_{1}$ and half with $_{2}$ .Both $_{1}$ and $_{2}$ have a thickness dd, but a cross-sectional area of $\frac{A}{2}$ , as shown in the diagram above.In terms of $C$ , what is the new capacitance of this modified system?

A

\frac{C}{2}

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B

C

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C

\frac{3 C}{2}

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D

2 C

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E

4 C

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Solution

The correct option is C $\frac{3 C}{2}$
Case I:
$C_{1} = \frac{K_{1} ε_{o} A}{d} ⟶ (i)$
Case II:
$C^{'} = C_{1} + C_{2}$
$C^{'} = \frac{K_{1} ε_{o} A}{2 d} + \frac{K_{2} ε_{o} A}{2 d}$
$= \frac{K_{1} ε_{o} A}{2 d} + \frac{2 K_{1} ε_{o} A}{2 d}$
$= \frac{3 K_{1} ε_{o} A}{2 d} ⟶ (i i)$
$= \frac{3 C}{2}$

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