A parallel plate capacitor of area A and separation d is charged to potential difference V and removed

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A parallel plate capacitor of area A and separation d is charged to potential difference V and removed from the charging source. A dielectric slab of constant K = 2, thickness d and area A/2  is inserted, as shown in the figure. Let σ1 be free charge density at the conductor-dielectric surface and σ2 be the charge density at the conductor-vacuum surface. (A) The electric field have the same value inside the dielectric as in the free space between the plates.

(B) The ratio σ12

(C) The new capacitance is 3∈0A/2d

(D) The new potential difference is 2/3 V

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Correct option (A,B,C,D)

Explanation:

Potential for each plate remain same over whole area. If potential difference between them is, say V' then V' = Ed i.e. E is also same inside the plates. To keep E same, free charge density is changed i.e. charge redistributes itself. To find new capacitance, two capacitors can be taken as connected in parallel. Then By Q = CV, as Q remains unchanged V is changed to 2/3 V.

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