The two metallic plates of radius `r` are placed at a distance `d` apart and its capacity is `C`. If a plate of radius `r//2` and thickness `d` of dielectric constant `6` is placed between the plates of the condenser, then its capacity will be

Options

A. `7 C//3`
B. `3C//7`
C. `7C//3`
D. `9C//4`

Answer

D

Explanation

Area of the given metallic plate `A = pir^(2)`
Area of the dielectric plate `A' = pi ((r )/(2))^(2) = (A)/(4)`
Uncovered area of the metallic plates `A'' = A - A'`
`= A - (A)/(4) = (3A)/(4)`
The given situation is equivalent to a parallel combination of two capacitor. One capacitor `(C')` is filled with a dielectric medium `(K =6)` having area `(A)/(4)` while the other capacitor (`C')` is air filled having area `(3A)/(4)`
Hence `E_(eq) = C' + C'' = (K epsilon_(0)(A//4))/(d) + (epsilon_(0) (3A//4))/(d)`
`= (epsilon_(0)A)/(d) ((K)/(4)+(3)/(4)) = (epsilon_(0)A)/(d) ((6)/(4)+(3)/(4)) = (9)/(4) C` (`:'C = (epsilon_(0)A)/(d))`