A parallel-plate capacitor of area A and separation d is charged to a potential difference V and is then removed from the charging source. A dielectric slab of constant k = 2, thickness d, and area ½A is inserted as shown in Figure. Let ?1 be the free charge density at the conductor???dielectric surface and ?2 be the free charge density at the conductor???air surface.

(a) Why must the electric field have the same value inside the dielectric as in the free space between the plates?

(b) Show that ?1 = 2?2.

(c) Show that the new capacitance is 3?0A/2d and that the new potential difference is V

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