The capacitance of a parallel plate condenser is `C_(0)` If a dielectric of relative permittivity `epsilon_(r)` and thickness equal to one fourth the plate separation is placed between the plates, then its capacity becomes C. The value of `(C)/(C_(0))` will be -

Options

A. `(5epsi_(r))/(4epsi_(r)+1)`
B. `(4epsi_(r))/(3epsi_(r)+1)`
C. `(3epsi_(r))/(2epsi_(r)+1)`
D. `(2epsi_(r))/(2epsi_(r)+1)`

Answer

B

Explanation

`C_(o)=(epsi_(o)A)/(d),C=(epsi_(o)A)/(d-(d)/(4)+(d)/(4epsi_(r)))=(epsi_(0)A)/((3d)/(4)-(d)/(4epsi_(r)))`
`=((4epsi_(r))/(3epsi_(r))+1)C_(0)`