A slab of material of dielectric constant $K$ has the same area as that of the plates of a parallel plate capacitor but has the thickness $\dfrac{2d}{3}$, where $d$ is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.

$C = \dfrac{3K\varepsilon_0 A}{d(2 + K)}$.

CBSE Class 12 · 2013 · Delhi · CapacitanceShort Answer

A slab of material of dielectric constant

K

has the same area as that of the plates of a parallel plate capacitor but has the thickness

\frac{2 d}{3}

, where

d

is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.

Answer: $C = \frac{3 K ε _{0} A}{d ( 2 + K )}$ .

The slab (thickness

\frac{2 d}{3}

, constant

K

) and the remaining air gap (thickness

\frac{d}{3}

) are in series:

\frac{1}{C} = \frac{2 d /3}{K ε _{0} A} + \frac{d /3}{ε _{0} A}

\frac{1}{C} = \frac{d}{3 ε _{0} A} (\frac{2}{K} + 1) = \frac{d ( 2 + K )}{3 K ε _{0} A}

C = \frac{3 K ε _{0} A}{d ( 2 + K )}

CBSE Class 12 Board · 2013 · Delhi · Subjective