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  4. 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

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Q. 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

Electrostatic Potential and Capacitance

Solution:

$C_{ air }=\frac{\varepsilon_{0} A}{d} ;$ With dielectric slab,
$C'=\frac{\varepsilon_{0} A}{\left(d-t+\frac{t}{K}\right)}$
Given: $C'=\frac{4}{3} C \Rightarrow \frac{\varepsilon_{0} A}{\left(d-t+\frac{t}{K}\right)}$
$=\frac{4}{3} \times \frac{\varepsilon_{0} A}{d}$
$\Rightarrow K=\frac{4 t}{4 t-d}$
$=\frac{4(d / 2)}{4[(d / 2)-d]}=2$