A parallel plate capacitor with air between the plates has a capacitan

1.	A parallel plate capacitor with air between the plates has a capacitance of 8pF (1 pF=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 s?
Answer» 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.

A parallel plate capacitor with air between the plates has a capacitance of 8pF (1 pF=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 s?

Answer»

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.