A parallel plate capacitor with air as the dielectric has capacitance C. A slab of dielectric constant K and having the same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be

$(K + 3) \frac{C}{4}$

$(K + 2) \frac{C}{4}$

$(K + 1) \frac{C}{4}$

$\frac{KC}{4}$

Solution

$(K + 3) \frac{C}{4}$

The condenser with air as the dielectric has capacitance

$C_{1} = \frac{ε_{o}}{d} (\frac{3 A}{4})$

C₁ = $\frac{3 ε_{o} A}{4 d}$

Similarly, the condenser with K as the dielectric constant has capacitance

$C_{2} = \frac{ε_{o} K}{d} (\frac{A}{4})$

$C_{2} = \frac{ε_{o} AK}{4 d}$

Since C₁ and C₂ are in parallel

C_net = C₁ + C₂

= $\frac{3 ε_{o} A}{4} [\frac{3}{4} + \frac{K}{4}]$

C_net = $\frac{C}{4} (K + 3)$

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