Question

Capacitance of a parallel plate capacitor becomes $4/3$ times its original value if a dielectric slab of thickness $t=d/2$ is inserted between the plates ( $d$ is the separation between the plates). The dielectric constant of the slab is

• A. 8
• B. 4
• C. 6
• D. 2

Answer 1

Answer: (D)

$C_{air}=\frac{{\epsilon }_{0}A}{d};$ With dielectric slab,
$C^{\prime }=\frac{{\epsilon }_{0}A}{\left(d-t+\frac{t}{K}\right)}$
Given: $C^{\prime }=\frac{4}{3}C\Rightarrow \frac{{\epsilon }_{0}A}{\left(d-t+\frac{t}{K}\right)}$
$=\frac{4}{3}\times \frac{{\epsilon }_{0}A}{d}$
$\Rightarrow K=\frac{4t}{4t-d}$
$=\frac{4(d/2)}{4[(d/2)-d]}=2$