Correct Answer - `C = (in_(0)A (3 d^(2) + 6 bd + 2b^(2))//3d (d + b) (d + 2b)`
As total area of flat plate is `A` and it is divided into there equal parts, forming three capacitors, therefore, area of each capacitor `= A//3` Starting from left end, fig.
`C_(1) = (in_(0) A//3)/(d) , C_(2) = (in_(0) A//3)/((d + b)), C_(3) = (in_(0) A//3)/((d + 2b))`
As the three capacitors are joined in parallel, therefore, total capacity of the assembly
`C_(p) = C_(1) + C_(2) + C_(3)`
`= (in_(0) A//3)/(d) + (in_(0) A//3)/(d + b) + (in_(0) A//3)/(d + 2b)`
`c_(P) = (in_(0) A)/(3)`
`xx [((d + b) (d + 2b) + d(d + 2b) + d(d + b))/(d(d + b) (d + 2b))]`
`= (in_(0)A (3 d^(2) + 6 bd + 2b^(2))//3d (d + b) (d + 2b)`