When a dielectric slab of thickness `d//2` is inserted between the plates of a capacitor (with d as the separation between the plates), the capacitance of a capacitor becomes `3//4` times the original value. Find the dielectric constant of the slab.

Options

A. 4
B. 8
C. 2
D. 6

Answer

C

Explanation

`E_(1)=(1)/(2)(Q^(2))/(C_(0)//2)=C_(0)V_(0)^(2), E_(2)=(1)/(2)(C_(0))/(2).V_(0)^(2)=(1)/(4)C_(0)V_(0)^(2), (E_(1))/(E_(2))=(4)/(1)`