The capacitance of capacitor with air as dielectric is given by
C = \(\frac{\varepsilon_0A}{d}\)
Given C = 8pF = 8 × 10-12F……(1)
If C is new capacitance when d1 = \(\frac{d}{2}\)and space is filled with a substance of dielectric constant k = 6. Then
C1 = \(\frac{\varepsilon_0kA}{d_1}=\frac{\varepsilon_0kA}{d/2}\)
or C1 = \(\frac{2k\varepsilon_0A}{d}\)
Using Eq.(1)
C1 = 12 × 8 × 10-12 F
or C1 = 96pF.