A parallel plate capacitor with air as a dielectric has capacitance C. A slab of dielectric constant K, having 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

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

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

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

$\frac{K C}{4}$

Solution

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

The two condensers with Kand with air are in parallel.

With air

C₁ = $\frac{ε_{0}}{d} (\frac{3 A}{4})$

C₁ = $\frac{3 ε_{0} A}{4 d}$

With medium

C₂ = $\frac{ε_{0} K}{d}$ $(\frac{A}{4})$

C₂ = $\frac{ε_{0} AK}{4 d}$

∴ C' = C₁ + C₂

= $\frac{3 ε_{0} A}{4 d} + \frac{ε_{0} AK}{4 d}$

C' = $\frac{ε_{0} A}{d} [\frac{3}{4} + \frac{K}{4}]$

⇒ C' = $\frac{C}{4} (K + 3) [∵ C = \frac{A ε_{0}}{d}]$

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