Find the capacitance between A and B if two dielectric slabs (each of thickness d) of dielectric constants `K_(1)` and `K_(2)` and areas `K_(1)` and `K_(2)` and inserted between the plates of a parallel plate capacitor of plate area `A`.
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Explanation

` (##MOT_CON_JEE_PHY_C25_SLV_048_S01.png" width="80%">
` C_1 =(A_1K_1in_0)/( d) , C_2=(A_2K_2in_0)/( d) `
` E_1 =(V)/(d) =(sigma _1)/(K_1in _0) ,E_2 =(V)/(d) = ( sigma _2)/( K_2in _0) `
` (##MOT_CON_JEE_PHY_C25_SLV_048_S02.png" width="80%">
` sigma _1 =(K_1in_0V)/(d) " "sigma _2 = (K_2in_0V)/( d) `
`C= ( Q_1+Q_2)/( V) = (sigma _1A_1 +sigma _2A_2)/(V) =(K_1in_0A_1)/( d) +( K_2in _0A_2 )/(d) `
` (##MOT_CON_JEE_PHY_C25_SLV_048_S03.png" width="80%">
The combination is equivalent to : ` C=C_1 +C_2 `