Calculate the ratio of capacitance of two capacitors of same dimension of same dimensions but of different values K and `(K)/(4)` arranged in two ways as shown in Fig. (i) and (ii).

Options

A. `5:2`
B. `25:16`
C. `5:4`
D. `2:5`

Answer

B

Explanation

Here, `C_(1) = (epsilon_(0)K_(1)A)/(2d) + (epsilon_(0)K_(2)A)/(2d)`
`= (epsilon_(0)KA)/(2d) + (epsilon_(0)KA)/(8d)" "{:("Given", K_(1) = K),("and" K_(2) = (K)/(4)):}`
`=(epsilon_(0)KA)/(d)((1)/(2) + (1)/(8)) = (epsilon_(0)KA)/(d)((4 + 1)/(8))`
`rArr C_(1) = (5epsilon_(0)KA)/(8d)" .....(i)"`
and `C_(2) = (epsilon_(0)A)/((d)/(2K_(1) + (d)/(2K_(2))) = (epsilon_(0)A)/((d)/(2K) + (d)/((2K)/(4))) = ((2K)epsilon_(0)A)/(5d)" ......(ii)"`
Diving Eq. (i) by Eq. (ii), we get
`(C_(1))/(C_(2)) = ((5epsilon_(0)KA)/(8d))/((2Kepsilon_(0)A)/(5d)) = (5xx5)/(2xx8) = (25)/(16)`