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?

Answer 1

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

\(C = \frac{kε°A}{d} = \frac{ε°A}{d} …………………. (1)\)

Where, A = Area of each plate ε_° = Permittivity of free space If distance between the plates is reduced to half, then new distance, d1 = d/2 Dielectric constant of the substance filled in between the plates, k1= 6 Hence, capacitance of the capacitor becomes.

\(C_1 =\frac{ k1ε°A}d_1 =\frac{6ε°A/d}{2}=\frac{12ε°A}{d}………………….( 2)\)

Taking ratios of equations (1) and (2), we obtain

C₁ = 2 × 6 C = 12 C = 12 × 8 pF = 96 pF

Therefore, the capacitance between the plates is 96 pF.

Answer 2

 \(C = 8pF = 8 × 10^{-12}\ F\)
\(C=\frac {ε_0A}{d}\)

\(8\times 10^{-12}=\frac {ε_0A}{d}\)

New capacitance \(C'=\frac {ε_0KA}{d'}\)
Where \(K=6\) and \(d'=\frac d2\)

\(C'=\frac {ε_0×6×A}{\frac d2}\)

\(C'=\frac {12×ε_0×A}{d}\)
\(C'=12×8×10^{-12}\)
\(C'=96×10^{-12}\ F\)
\(C'=96\ pF\)

So, the answer is \(96\ pF\).