Two metal plates form a parallel plate capacitor. The distance between the plates is d. A metal sheet of thickness d/2 and of the same area is indroduced between the plates. What is the ratio of the capacitances in the two cases?

Options

A. `4:1`
B. `2:1`
C. `3:1`
D. `5:1`

Answer

B

Explanation

The capacitance of the air filled parallel plate capacitor is given by
`C = (epsilon_(0)A)/(d)` …(i)
When a slab of dielectric constant K, and thickness t is introduced in between the plates of the capacitor, its new capacitance is given by,
`C. = (epsilon_(0)A)/(d-t(1-(1)/(K)))`
Since a metal sheet of thickness d/2 is introduced, hence here `t = d//2, K oo` (for metals)
or `(1)/(K) = 0 :. C. = (epsilon_(0)A)/(d-(d)/(2)) = (2 epsilon_(0)A)/(d)` ...(ii)
Hence, from eqs. (i) and (ii), we get
`:. (C.)/(C ) = ((2epsilon_(0)A)/(d))/((epsilon_(0)A)/(d)) = (2)/(1) = 2:1`