A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constants 1 and 2, as shown in figures. The distance between the plates is d and area of each plate is A. If capacitance in first configuration and second configuration are C1 and C2 respectively, then C1C2 is: First Configuration [figure] Second Configuration [figure]
- 0( 1 + 2)2
- 1 22( 1 + 2)2
- 4 1 2( 1 + 2)2
- 1 2 1 + 2
Answer
(3) 4 1 2( 1 + 2)2
Solution
First configuration: two slabs, each of thickness d2 and full area A, lie one above the other between the plates series.
C' = 1 0 Ad/2 = 2 1 0 Ad, C'' = 2 2 0 Ad
C1 = C'C''C' + C''
= 2 0 Ad 1 2 1 + 2
Second configuration: two slabs side by side, each of area A2 and thickness d parallel.
C2 = 1 0 (A/2)d + 2 0 (A/2)d
= 0 A( 1 + 2)2d
C1C2 = 2 1 2 1 + 2 2 1 + 2
= 4 1 2( 1 + 2)2
Answer: (3)