When a dielectric slab of the same area of cross-section and thickness equal to `2/3` of the separation is introduced between the plates of a parallel plate capacitor, its capacitance becomes 2.25 times the original value. The dielectric constant of the material of the slab is

Options

A. 1.5
B. 4.5
C. 5
D. 6

Answer

(d)

Explanation

Capacitance without dielectric slab,`C_(a) = A epsilon_(0)/d`
Capacitance with dielectric slab,
`C' = Aepsilon_(0)/d - t + 1/k`
` = Aepsilon_(0)/d - 2d/3 + 2d//3/k `
`= Aepsilon_(0)/d/3 + 2d/3k `
`rArr 2.25 C_(a) = 3C_(a)/(1 + 2/k)`
`1 + 2/k = 3/2.25 `
`:. ` k =6