The expression for the capacity of a capacitor formed by placing a compound dieletric between the plates of a parallel plate capacitor, as shown in figure , will be (Given area of plate = A)

Options

A. `(epsilon_(0)A)/((d_(1)/(K_(1))+(d_(2))/(K_(2))+(d_(3))/(K_(3))))`
B. `(epsilon_(0)A)/((d_(1)/(K_(1))+(d_(2))/(K_(2))+(d_(3))/(K_(3))))`
C. `(epsilon_(0)A(K_(1)K_(2)K_(3)))/((d_(1)d_(2)d_(3)))`
D. `epsilon_(0)((AK_(1))/(d_(1))+(AK_(2))/(d_(2))+(AK_(3))/(d_(3)))`

Answer

a

Explanation

Capacity of parallel plate condenser `=(epsilon_(0)KA)/(d)`
`C=(epsilon_(0)A)/(d//K)`. Three capacitances are `C_(1),C_(2)andC_(3)`.
`thereforeC_(1)=(epsilon_(0)A)/(d_(1)//K_(1)),C_(2)=(epsilon_(0)A)/(d_(2)//K_(2)),C_(3)=(epsilon_(0)A)/(d_(3)//K_(3))`
`C_(1),C_(2)andC_(3)` are in series.
`therefore(1)/(C)=(1)/(C_(1))+(1)/(C_(2))+(1)/(C_(3))=(d_(1)//K_(1))/(epsilon_(0)A)+(d_(2)//K_(2))/(epsilon_(0)A)+(d_(3)//d_(3))/(epsilon_(0)A)`
`rArr(1)/(C)=(1)/(epsilon_(0)A)((d_(1))/(K_(1))+(d_(2))/(K_(2))+(d_(3))/(K_(3)))`
`rArrC((d_(1))/(K_(1))+(d_(2))/(K_(2))+(d_(3))/(K_(3)))=epsilon_(0)ArArrC=(epsilon_(0))/(((d_(1))/(K_(1))+(d_(2))/(K_(2))+(d_(3))/(K_(3))))`