Question

When a dielectric slab of thickness 6cm is introduced between the plates of parallel plate condenser, it is found that the distance between the plates has to be increased by 4cm to restore the capacity to original value. The dielectric constant of the slab is

• A. 1.5
• B. 2/3
• C. 3
• D. 4

Answer 1

The correct option is C 3

Increase in the distance between the plates = (x) = 4 cm

Thickness of Dielectric slab= t=6 cm

Capacitance is given by $C=\frac{{ϵ}_{0}A}{d-t+\frac{t}{K}}$

$\Rightarrow$ To restore the original value $x=t(1-\frac{1}{K})\Rightarrow K=\left(\frac{t}{t-x}\right)\Rightarrow K=\frac{6}{6-4}=\frac{6}{2}=3$