Question

An air-filled parallel plate capacitor has a capacity $2pF$. The separation between the plates is doubled and the interspace is filled with wax, If the capacity is increased to 6 pF, the dielectric constant of the wax is :

• A. $2$
• B. $4$
• C. $3$
• D. $6$

Answer 1

The correct option is D $6$

when Air filled , capacitance $=C=2pF$
As C is inversely proportional to the separation so it is double the capacitance will be $C/2=$
If dielectric constant is $k$, after filling wax the capacitance becomes $C^{\prime }=k\frac{C}{2}$
now $k=\frac{C^{\prime }}{C/2}=\frac{6}{2/2}=6$