Question

A parallel-plate air capacitor has a capacitance of $4\mu F$ what be its new capacitance if a slab of dielectric constant 5 is introduced filling the entire space between the two plates?

Answer 1

The capacitance is given as,

$C=\frac{\epsilon A}{d}$

$4=\frac{\epsilon A}{d}$

The new capacitance is given as,

$C_n=\frac{5\epsilon A}{d}$

$C_n=5\times 4$

$C_n=20\mu F$