1.

What is the capacity having a plate area A of the figure given below?

Answer»

Solution :Capacity without dielectric `=C_(0)=(epsi_(0) A)/(2d)`
Capacity with dielectric , `C=(epsi_(0) KA)/(2d)`
Here capacity with dielectic constant `K_(1)=C_(1)=(epsi_(0) K_(1) (A/2))/(2d) =(epsi_(0)K_(1)A)/(4d)`
Similarly with `K_(2), C_(2)=(epsi_(0) K_(2) (A/2))/(d)=(epsi_(0 K_(2)A))/(2d) and" with "K_(3), C_(3)=(epsi_(0)K_(3)A)/(2d)`
`C_(2) and C_(3)" in SERIES "C.=(C_(2)C_(3))/(C_(2)+C_(3))=((epsi_(0)K_(2)A)/(2d).(epsi_(0)K_(3)A)/(2d))/((epsi_(0)K_(2)A)/(2d)+(epsi_(0)K_(3)A)/(2d))`
`=(epsi_(0)K_(2)A epsi_(0)K_(3)A)/(4d^(2)) xx (2d)/(epsi_(0) A(K_(2)+K_(3))) =(epsi_(0) A.K_(2)K_(3))/(2d(K_(2)+K_(3))`
`C. and C_(1)` are parallel
`C=C. +C_(1)= (epsi_(0) AK_(2)K_(3))/(2d(K_(2)+K_(3))+(epsi_(0)K_(1)A)/(4d))=(epsi_(0)A)/(4d) [K_(1)+(2K_(2)K_(3))/((K_(2)+K_(3)))]`


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