The two metallic plates of radius r are placed at a distance d apart and its capacity is C. If a plate of radius r / 2 and thickness d of dielectric constant 6 is placed between the plates of the condenser, then its capacity will be

Area of the given metallic plate A=πr2Area of the dielectric plate A′=πr22=A4Uncovered area of the metallic plates A′′=A−A′=A−A4=3A4The given situation is equivalent to a parallel combination of two capacitor. One capacitor (C') is filled with a dielectric medium (K = 6) having area A4 while the other capacitor (C'') is air filled having area 3A4Hence Ceq=C'+C"=Kε0 (A/4)d+ε0(3A/4)d=ε0AdK4+34=ε0Ad64+34=94C ∵C=ε0Ad