A parallel plate capacitor with air between the plates has a capacitance of 6 PF. What will be the capacitance if the distance between the plates is reduced to half and the space between them is filled with a substance of dielectric constant 5?

60 pF

The correct answer is Option (2) → 60 pF

Capacitance of a parallel plate conductor is,

$C=\frac{ε_0A}{d}$

for initial conductor with air between the plates -

$C_1=\frac{ε_0A}{d}=6pF$

After modifying the conditions -

$C_2=\frac{kε_0A}{d/2}=2k\left(\frac{ε_0A}{d}\right)$

$=2×5×6=60pF$