Question

When a dielectric slab of thickness 6 cm is introduced between the plates of parallel plate condenser it is found that the distance between the plates has to he increased by 4 cm to restore to 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 $3$

Let initial capacitance be $C$

$C=\frac{{\epsilon }_{o}A}{d}$ $Aandd$ are in $cm$

Final capacitance is same as initial capacitance.

$\frac{d}{{\epsilon }_{o}A}=\frac{d+4-6}{{\epsilon }_{o}A}+\frac{6}{K{\epsilon }_{o}A}$

Solving, we get:

$K=3$