Two parallel plate capacitors have the same separation d = 8.85 × 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 K = 9 has dimensions such that it can exactly fill space between the plates of capacitor B. The slab is placed inside A as shown in figure
Fig. The battery is disconnected and then the dielectric slab is removed from A. The same dielectric slab is now placed inside B filling it completely. The two capacitors are then connected as shown in figure The work done by an external agency in removing the slab from A is 4.84 × 10 –5 J
Fig. The work done on an external agency in removing the slab from A is 4.84 × 10 –5 J
Text Solution
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The capacitance of A without dielectric is
C 0 = 
The energy present in A after removal of the slab is, therefore, U f = 
= 605 × 10 –7 J where q = CV = 2 × 10 –9 × 110 = 22 × 10 –8 coulomb. The increase in energy is, therefore, Δ U = 605 × 10 –7 – 121 × 10 –7 = 484 × 10 –7 J. This is the required work done. The option is correct. Now, the capacitance of A without dielectric is
C A = 0.4 × 10 –9 F
The capacitance of B with dielectric is
= 1.8 × 10 –9 F
The total capacitance of the system is
C eff = 0.4 × 10 –9 + 1.8 × 10 –9
= 2.2 × 10 –9 F
The charge on the system has already been obtained. It is q = CV = 22 × 10 –8 coulomb. So, the energy stored on the system is:
= 11 × 10 –6 J
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