A parallel plate air capacitor has a capacitance $C$ . When it is half filled with a dielectric $(K = 2)$ , the percentage increase in the capacitance will be)

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Solution

Given arrangement can be considered as a combination of two capacitors namely $C_{1} and C_{2}$ connected between $A and B$ as shown in the diagram.

We can calculate the capacitace of each one as:

$C_{1} = \frac{K (\frac{A}{2}) ϵ_{0}}{d} = 2 \times \frac{A ϵ_{0}}{2 d} = \frac{A ϵ_{0}}{d} = C$

$C_{2} = \frac{(\frac{A}{2}) ϵ_{0}}{d} = \frac{A ϵ_{0}}{2 d} = \frac{C}{2}$

Since, $C_{1}$ and $C_{2}$ are in parallel (check polarity)
So,
$C_{e q} = C_{1} + C_{2} = 3 C / 2$

Initial capacitance $= C$

Final capacitance $= \frac{3 C}{2}$

Thus, $%$ change is $: (\frac{3 C / 2}{C} - 1) \times 100 = 50 %$
Hence, the correct option is $(b)$ .

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