Suppose n conducting face to face, and the distance between two successive plates is d. Each plate is half of the area of the previous one. If area of first plate is A
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(i) If the area of the first plate is A, find the equyivalent capacitance of the system.
(ii) If a cell of emf V is connected as shown in and a dielectric of dielectric constant k is inserted between the first and second plates, dind the charge on the nth plate.

Answer

i. `(epsilon_(0)A)/(d(2^(n)-2))` ii. `-(Aepsilon_(0))/(2d[2^(n-1)-2+1/k])V`

Explanation

(i) `(1)/(C_(eq))=(1)/((epsi_(0)A)/(2d))+(1)/((epsi_(0)A)/(2^(2)d))+....+(1)/((epsi_(0)A)/(2^(n-1)d))=(2(2^(n-1)-1)d)/(epsi_(0)A)`
or `C_(eq) = (epsi_(0)A)/(d(2^(n) - 2))`
If dielectric is placed,
`1/C_(eq)=1/((epsilon_(0)kA)/(2d))+1/((epsilon_(0)A)/(2^(2)d))+…+1/((epsilon_(0)A)/(2^(n-1)d))`
or `C_(eq)=(Aepsilon_(0))/(2d[2^(n-1)-2+1/k])`
Charge on capacitor is `C_(eq)V`
`Q=`Charg e on nth plate`= -q =-C_(eq)V`.