`n` conducting plates are placed face to face. Distance between two consecutive plates is `d`.
Area of plates is `A,(A)/(2),(A)/(4),(A)/(8)....((1)/(2^(n-1)))A`
A dielectric slab of dielectric constant `k` is inserted between the first and second plates and the assembly is charged by a battery of `emf xi`. Find the charge stored in the assembly.
.

Options

A. `(epsilon_(0) A xi)/(2d[(1)/(k)+2^(n-1)-2])`
B. `(epsilon_(0) A xi)/(2d[(1)/(k)+2^(n-1)-1])`
C. `(epsilon_(0) A xi)/(d[(1)/(k)+2^(n-1)-2])`
D. `(epsilon_(0) A xi)/(3d[(1)/(k)+2^(n-1)-2])`

Answer

A

Explanation

`(n-1)` capacitors are in series, So,
`(1)/(C)=(1)/((epsilon_(0)A//2)/(d))+(1)/((epsilon_(0) A//4)/(d))+...+(1)/((epsilon_(0) A//2^(n-1))/(d)s)`
`C_(eq)=(epsilon_(0)A)/(2d[(1)/(k)+2^(n-1)-2])`
` :. q=C_(eq) xi rArr 1=(epsilon_(0) A xi)/(2d[(1)/(k)+2^(n-1)-2])`
`V_(1) = 30 V and V_(2) = 30 V` (Conceptual).