Two parallel plate capacitors A and B have the same separation d = 8.85 x 10 -4 m between the plates. The plate areas of A and B are 0.04 m 2 and 0.02 m 2 respectively. A slab of dielectric constant (relative permittivity K = 9) has dimensions such that it can exactly fill the space between the plates of capacitor B.
(i) The dielectric slab is placed inside A as shown in the figure
Text Solution
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[(i) 2nF, 12.1 µJ, (ii) 48.4µJ, (iii) 11µJ]
Sol.
C A = 
(i) C A = 
= 2 × 10 –9 F = 2nF
U A = 
C A V 2 = 
× 2 × 10 –9 × (110) 2 = 121 × 10 –7 J = 12.1µJ Ans
(ii) W = 
C air = 
= 0.4nF
Q = C A V = 2 × 10 –9 × 110 = 0.22µC
U f = 
= 60.5µJ,
U i = 12.1 µJ
⇒ W = 48.4µJ Ans
(iii) C A = 0.4 nF, Q A = 0.22µC, C B = 
= 18 × 10 –10 F = 1.8 nF
common potential V = 
= 100V
U = 
C A V 2 + 
C B V 2 = 
(C A + C B ) V 2
= 
(0.4 + 1.8) × 10 –9 (100) 2 = 11µJ Ans
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