Two dielectric slabs of relative oremittivities
`epsilon_(r_(1))=2` and `epsilon_(r_(2))=3` and thickness `s=2d` are filled in between the grounde plattery each of area A If a battery of emf `epsilon` is connected to the middle conducting plate, dind the
a. capacitance of the system,
b. charge flowing through the battery,
c. energy stored in the system.
.

Answer

a. `(7epsilon_(0)A)/(2d)` b. `(7epsilon_(0)A)/(2d)` c. charge on plate (1), `-(2epsilon_(0)Aepsilon)/d`
Charge on palate (2), `(7epsilon_(0)Aepsilon)/(2d)`
Charge on plate (3), `(-3epsilon_(0)Aepsilon)/(2d)`
d. `(7epsilon_(0)Aepsilon)/(4d)`

Explanation

Equivalent circuit is two capacitors in parallel
.
`C_(1)=(epsilon_(0)epsilon_(r_(1))A)/d=(2epsilon_(0)A)/d` (i)
`C_(2) = (epsi_(0)epsi_(r_(2))A)/(d)=(3epsi_(0)A)/(2d)` (ii)
`C_(1)` and `C_(2)` are in parallel.
`C_(eq)=C_(1)+C_(2)+C_(2)=(7epsilon_(0)A)/(2d)`
Charge drawn through the battery is
`Q=C_(eq)epsilon=(7epsilon_(0)Aepsilon)/(2d)`
charge in `C_(1)` is `Q_(1)=C_(1)epsilon=(2epsilon_(0)Aepsilon)/d`
And charge in `C_(2)` is `Q_(2)=C_(2)epsilon=(3epsilon_(0)Aepsilon)/(2d)`
The charge on outer surfaces of plates will be zero.
Hence, charge on plate (1) is `q_(1)=-(2epsilon_(0)Aepsilon)/d`
Charge on plate `(2)` is `q_(2)=(2epsilon_(0)A epsilon)/d+(3epsilon_(0)A epsilon)/(2d)=(7epsilon_(0)Aepsilon)/(2d)`
Charge on `(3)` is `q_(3)=(-3epsilon_(0)Aepsilon)/(2d)`
Energy stored in system is `U=1/2C_(eq)epsilon^(2)=(7epsilonAepsilon^(2))/(4d)`.