Half of the space between the plates of a parallel plate capacitor is filled with a dielectric material of dielectric constant K . The ramaining half contains air as shown in the figure . The capacitor is now given a charge Q . Then

Options

A. electric field in the dielectric filled region is higher than that in the air - filled region.
B. on the two halves of the bottom plate the charge densities are unequal.
C. charge on the half of the top plate above the air filled part is `(Q)/(K+1)`.
D. capacitance of the capacitor shown above is `(1+K)(C_(0))/(2)`, where `C_(0)` is the capacitance of the same capacitor with dielectric removed.

Answer

d

Explanation

Let A be area of each plate and d is the distance between the plates.
The given capacitor is equivalent to two capacitors in parallel with capacitances
`C_(1)=(Kepsilon_(0)(A//2))/(d)=(Kepsilon_(0)A)/(2d)` … (i)
`C_(2)=(epsilon_(0)(A//2))/(d)=(epsilon_(0)A)/(2d)` ...(ii)

`thereforeC_(eq)=C_(1)+C_(2)=(Kepsilon_(0)A)/(2d)+(epsilon_(0)A)/(2d)`
`=(epsilon_(0)A)/(2d)(K+1)=(C_(0))/(2)=(K+1)`, Where `C_(0)=(epsilon_(0)A)/(d)`
`(Q_(1))/(Q_(2))=(C_(1))/(C_(2))=(K)/(1)` ...(iii) (Using (i) and (ii))
As surface charge density , `sigma=("Charge")/("Area")`
`therefore(sigma_(1))/(sigma_(2))=(Q_(1))/(Q_(2))=(K)/(1)` ...(iv) (Using (iii))
Total charge , `Q=Q_(1)+Q_(2)` ...(v)
From (iii) and (v) , we get
`Q_(1)=(KQ)/(K+1)andQ_(2)=(Q)/(K+1)`
Electric field in dielectric filled region,
`E_(1)=(sigma_(1))/(epsilon_(0)K)`
Electric field in air - filled region , `E_(2)=(sigma_(2))/(epsilon_(0))`
`therefore(E_(1))/(E_(2))=(sigma_(1))/(sigma_(2))xx(1)/(K)=Kxx(1)/(K)=1` (Using (iv))