1.

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 airas shown in the figure . The capacitor is now given a charge Q . Then

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electric field in the dielectric filled region is higher than that in the air - filled region.
on the two HALVES ofthe bottom PLATE the charge densities areunequal.
charge on the half of the top plate above the air filled part is `(Q)/(K+1)`.
capacitance of the capacitor shown above is `(1+K)(C_(0))/(2)`, where `C_(0)` is the capacitance of the samecapacitor with dielectric removed.

Solution :Let A be area of each plate and d is the distance between the plates.
The given capacitor is equivalent to twocapacitors 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))


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