Question

In the arrangement shown in figure dielectric constant `K_1=2` and `K_2=3`. If the capacitance across `P` and `Q` are `C_1` and `C_2` respectively, then `C_1//C_2` wil be (the gaps shown are negligible)
`
A. `1:1`
B. `2:3`
C. `9:5`
D. `25:24`

Answer 1

Correct Answer - D
`C_1=C_(RHS)+ C_(LHS)`
`=(K_2epsilon_0(A/2))/d=(K_1epsilon_0(A/2))/d`
`(epsilon_0A)/(2d)(K_1+K_2)=(5epsilon_0A)/(2d)`
`C_2=(epsilon_0A)/(d-d/2-d/2+(d/2)/K_1+(d/2)/K_2`
`=(2epsilon_0A)/d((K_1K_2)/(K_1+K_2))`
`=(12epsilon_0A)/(5d)`
`:. C_1/C_2=25/24`