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Question: Two parallel plate capacitors have the same separation d = 8.85 × 10<sup>–4</sup> m between the plat...

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 m2 and 0.02 m2 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 (1). A is then charged to a potential difference of 110 volt.

Fig.(1)

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 (2). Then:

A

The work done by an external agency in removing the slab from A is 4.84 × 10–5 J

B

The work done on an external agency in removing the slab from A is 4.84 × 10–5 J

C

The energy stored in the system in figure is 1.1 × 10–6 J

D

The energy stored in the system in figure is 11 × 10–6 J

Answer

The energy stored in the system in figure is 11 × 10–6 J

Explanation

Solution

The capacitance of A without dielectric is

C0 = 8.85×10−12×0.048.85×10−4\frac { 8.85 \times 10 ^ { - 12 } \times 0.04 } { 8.85 \times 10 ^ { - 4 } } = 0.4 × 10–9 F

The energy present in A after removal of the slab is, therefore, Uf = q22C\frac { q ^ { 2 } } { 2 C } = 605 × 10–7 J where q = CV = 2 × 10–9 × 110 = 22 × 10–8 coulomb. The increase in energy is, therefore, DU = 605 × 10–7 – 121 × 10–7 = 484 × 10–7 J. This is the required work done. The option (1) is correct. Now, the capacitance of A without dielectric is

CA = 0.4 × 10–9 F

The capacitance of B with dielectric is

CB = 8.85×10−12×9×0.028.85×10−4\frac { 8.85 \times 10 ^ { - 12 } \times 9 \times 0.02 } { 8.85 \times 10 ^ { - 4 } }

= 1.8 × 10–9 F

The total capacitance of the system is

Ceff = 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:

U = q22C\frac { q ^ { 2 } } { 2 C } = 12\frac { 1 } { 2 } ×(22×10−8)22.2×10−9\frac { \left( 22 \times 10 ^ { - 8 } \right) ^ { 2 } } { 2.2 \times 10 ^ { - 9 } }J

= 11 × 10–6 J

Therefore, the option (4) is also correct.