A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10^–12 F). 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?

Question

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10^–12 F). 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?

Philoid · Accepted Answer

Given: Capacitance of capacitor when medium between two plates is air, C = 8 pF = 8 × 10^–12 F

Now, the capacitance of a parallel plate capacitor is given as:

where, A be the area of each plate

d be the distance between the two plates of the parallel plate capacitor

k is the dielectric constant, (for air , k = 1)

Capacitance, C = 8 pF(Given)

⇒

Suppose that the capacitance of the capacitor becomes C’ when the distance between the plates is reduced to half (d’ = d/2) and the space between them is filled with a substance of dielectric constant K = 6.

⇒

From equation (i) and (ii),

C’ = 12 x 8 x 10^-12 = 96 x 10^-12 = 96 pF

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10–12 F). 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?

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10^–12 F). 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?