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Question

A dielectric slab fills the lower half of a parallel plate capacitor as shown in figure :
(Take plate are as A)

155243_2c1d9fc893f645b8affb28b037547f1c.png
  1. equivalent capacity of the system is ((ε0A/2d)(1+K))
  2. the net charge of the lower half of the left hand plate is 1/K times the charge on the upper half of the plate
  3. net charges on the lower and upper halves of the left hand plate are different
  4. net charge on the lower hand of the left hand plate is Kε0A2d×V

A
net charges on the lower and upper halves of the left hand plate are different
B
equivalent capacity of the system is ((ε0A/2d)(1+K))
C
the net charge of the lower half of the left hand plate is 1/K times the charge on the upper half of the plate
D
net charge on the lower hand of the left hand plate is Kε0A2d×V
Solution
Verified by Toppr

This system can be considered as two capacitors in parallel with
C1=ε0A2d and C2=Kε0A2d
Ceq=C1+C2=ε0A2d+Kε0A2d=ϵ0A2d×(1+K). Hence option A is correct
C1 is due to upper half of two plates, while C2 is due to lower half. As potential difference across two capacitors is the same, charges would be different as capacitors are different. Hence option C is correct,

Net charge on the lower half plate
Q2=C2V=Kε0A2dV.
Therefore option D is correct

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