Question

A parallel plate capacitor with air between the plates has a capacitance of 8 pF. Calculate 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. (ε_r = 6)

पर्याय

• 72 pF

• 81 pF

• 84 pF

• 96 pF

Answer 1

96 pF

Explanation:

Capacity of a parallel plate capacitor

C = `(ε_"r"ε_0"A")/"d"` ....(For air ε_r = i)

So, `(ε_0"A")/"d"` = 8 × 10⁻¹²

If `"d" → "d"/2` and ε_r → 6 then new capacitance

C' = `6 xx (ε_0"A")/("d"//2)`

= `12 (ε_0"A")/"d"`

= 12 × 8 pF

= 96 pF