Question

In a parallel-plate capacitor of capacitance C, a metal sheet is inserted between the plates, parallel to them. The thickness of the sheet is half of the separation between the plates. The capacitance now becomes 

(a) 4C 

(b) 2C 

(c) C/2 

(d) C/4

Answer 1

Correct Answer is: (b) 2C

Method 1 Before the metal sheet is inserted, C = ɛ₀A/d.

After the sheet is inserted, the system is equivalent to two capacitors in series, each of capacitance C' ɛ₀A(d/4) = 4C.

The equivalent capacity is now 2C.

Method 2 If a dielectric slab of dielectric constant K and thickness t is inserted in the air gap of a capacitor of plate separation d and plate area A, its capacitance becomes

C = ε₀A/d - t (1 - 1/k).

Here, the initial capacitance, C = ε₀A/d

For the metal sheet, t = d/2 , K = ∞.The new capacitance is