2.5 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?

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alok kumar singh | Contributor-Level 10

3 months ago

2.5 Capacitance between the parallel plates of the capacitor, C = 8 pF

Let us assume, initially, distance between the parallel plates was d and it was filled with air.

Dielectric constant of air, k = 1

Capacitance C is given by the formula,

C = $\frac{k ε_{0} A}{d}$ = $\frac{ε_{0} A}{d}$ , since k = 1…………………….(1)

Where A = area of each plate

$ε_{0}$ = permittivity of free space = 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

d = distance between two plates

If the distance between two plates is reduced to half, then distance between two plates $d_{1}$ = $\frac{d}{2}$

Dielectric constant of a new substance, $k_{1}$ = 6

Then the resistanc

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2.5 Capacitance between the parallel plates of the capacitor, C = 8 pF

Let us assume, initially, distance between the parallel plates was d and it was filled with air.

Dielectric constant of air, k = 1

Capacitance C is given by the formula,

C = $\frac{k ε_{0} A}{d}$ = $\frac{ε_{0} A}{d}$ , since k = 1…………………….(1)

Where A = area of each plate

$ε_{0}$ = permittivity of free space = 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

d = distance between two plates

If the distance between two plates is reduced to half, then distance between two plates $d_{1}$ = $\frac{d}{2}$

Dielectric constant of a new substance, $k_{1}$ = 6

Then the resistance between two plates, $C_{1}$ = $\frac{6 ε_{0} A}{d / 2}$ = $\frac{12 ε_{0} A}{d}$ ……………(2)

From equation (1) and (2) we get

$C_{1} = 12 C$ = 12 $\times 8$ pF = 96 pF

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