Question

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (lpF = 10^-12F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

Answer 1

C = 8pF,
C = \(\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}\)
C' = \(\frac{\varepsilon_{\mathrm{r}} \varepsilon_{0} \mathrm{~A}}{\frac{\mathrm{d}}{2}}\)
= 2ε_r
∴ C' = 2 x 6 x 8 x 10^-12 F = 96 pF