Question

A slab of dielectric constant $K$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4}d$, where $d$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be : (Given $C_0=$ capacitance of capacitor with air as medium between plates.)

• A. $\frac{4K{C}_0}{3+K}$
• B. $\frac{3K{C}_0}{3+K}$
• C. $\frac{3+K}{4K{C}_0}$
• D. $\frac{K}{4+K}$

Answer 1

Answer: (A)

$x+y+\frac{3d}{4}=d$
$x+y=\frac{d}{4}$
$\frac{A{\in }_0}{d}=C_0$
$\Delta V=Ex+\frac{E}{k}\times \frac{3d}{4}+Ey$
$=\frac{3Ed}{4k}+E(x+y)$
$\Delta V=E\left[\frac{3d}{4k}+\frac{d}{4}\right]$
$\Delta V=\frac{\sigma }{{ϵ}_0}\left[\frac{3d+dk}{4k}\right]=\frac{Qd}{A{\in }_0}\left[\frac{3+k}{4k}\right]$
$\frac{Q}{\Delta V}=C=\frac{A{\in }_0}{d}\left[\frac{4k}{3+k}\right]=\frac{4k{C}_0}{k+3}$