A parallel plate capacitor with air between the plates has a capacitance of 9 pF. What will be the capacitance if the distance between the plates is reduced to half and the space between them is filled by a substance of dielectric constant 7?

126 pF

The correct answer is Option (4) → 126 pF

The capacitance is given as,

$C=\frac{k.ε_0.A}{d_{new}}$

Since, $d_{new}=\frac{d_{new}}{2}$, this is

$C_{new}=\frac{k.ε_0.A}{\frac{d_{new}}{2}}=2×k×\frac{ε_0A}{d_{initial}}$

$=2×7×9$

$=126pF$