Question

The separation between the plates of the parallel plate capacitor is d and the area of each plate is A. When a slab of material of dielectric constant k and thickness t(t < d) is introduced between the plates, its capacitance becomes ______.

(ε₀ = permittivity of free space)

पर्याय

• `(Aε_0)/(2d - t[1 - 1/k])`

• `(Aε_0)/(d + t[1 + 1/k])`

• `(Aε_0)/(d - t[1 - 1/k])`

• `(Aε_0)/(d - t[1 + 1/k])`

Answer 1

The separation between the plates of the parallel plate capacitor is d and the area of each plate is A. When a slab of material of dielectric constant k and thickness t(t < d) is introduced between the plates, its capacitance becomes `underlinebb((Aε_0)/(d - t[1 - 1/k]))`.

Explanation:

The given situation can be shown as,

The capacitance of a parallel plate capacitor is given by,

C = `(kε_0A)/d`

The above circuit can be considered to be a combination of two series capacitors as,

 

∴ `1/C = 1/C_1 + 1/C_2 = ((d - t))/(ε_0A) + t/(kε_0A)`

= `(k(d - t) + t)/(kε_0A) = (d - t + t/k)/(ε_0A) = (d - t(1 - 1/k))/(ε_0A)`

⇒ C = `(Aε_0)/(d - t(1 - 1/k))`