Two parallel plate capacitors A and B have the same separation d=8.85 × 10-4m between the plates. The plate areas of A and B are 0.04m2 and 0.02m2 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. (a) The dielectric slab is placed inside A as shown in figure (i) A is then charged to a potential difference of 110 V. Calculate the capacitance of A and the energy stored in it (b) The battery is disconnected and then the dielectric slab is removed from A. Find the work done by the external agency in removing the slab from A. (c) The same dielectric slab is now placed inside B, filling it completely. The two capacitors A and B are then connected as shown in figure (iii). Calculate the energy stored in the system.

Question

Two parallel plate capacitors A and B have the same separation d=8.85 × 10-4m between the plates. The plate areas of A and B are 0.04m2 and 0.02m2 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. (a) The dielectric slab is placed inside A as shown in figure (i) A is then charged to a potential difference of 110 V. Calculate the capacitance of A and the energy stored in it (b) The battery is disconnected and then the dielectric slab is removed from A. Find the work done by the external agency in removing the slab from A. (c) The same dielectric slab is now placed inside B, filling it completely. The two capacitors A and B are then connected as shown in figure (iii). Calculate the energy stored in the system. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(a) Capacitor A is a combination of two capacitors

C_{K} an d

C_{0}

in parallel. Hence,

C_{A} = C_{K} + C_{0} = \frac{K ε _{0} A}{d} + \frac{ε _{0} A}{d} = (K + 1) \frac{ε _{0} A}{d}

Here,

A = 0.02 m^{2} .

Substituting the values, we have

C_{A} = (9 + 1) \frac{8.85 \times 1 0 ^{- 12} ( 0.02 )}{( 8.85 \times 1 0 ^{- 4} )}

C_{A} = 2.0 \times 1 0^{- 9} F

Energy stored in capacitor A, when connected with a
110 V battery is

U_{A} = \frac{1}{2} C_{A} V^{2} = \frac{1}{2} (2 \times 1 0^{- 9}) (110)^{2}

U_{A} = 1.21 \times 1 0^{- 5} J

(b) Charge stored in the capacitor

q_{A} = C_{A} V = (2.0 \times 1 0^{- 9}) (110)

q_{A} = 2.2 \times 1 0^{- 7} C

Now, this charge remains constant even after battery is
disconnected. But when the slab is removed, capacitance
of A will get reduced. Let it be

C_{A}^{'}

C_{A}^{'} = \frac{ε _{0} ( 2 A )}{d} = \frac{( 8.85 \times 1 0 ^{- 12} ) ( 0.04 )}{8.85 \times 1 0 ^{- 4}}

C_{A}^{'} = 0.4 \times 1 0^{- 9} F

Energy stored in this case would be

U_{A}^{'} = \frac{1 ( q _{A} ) ^{2}}{2 C _{A}^{'}} = \frac{1 ( 2.2 \times 1 0 ^{- 7} ) ^{2}}{2 ( 0.4 \times 1 0 ^{- 9} )}

U_{A}^{'} = 6.05 \times 1 0^{- 5} J > U_{A}

Therefore, work done to remove the slab would be

W = U_{A}^{'} - U_{A} = (6.05 - 1.21) \times 1 0^{- 5} J

or

W = 4.84 \times 1 0^{- 5} J

(c) Capacity of B when filled with dielectric is

C_{B} = \frac{K ε _{0} A}{d} = \frac{( 9 ) ( 8.85 \times 1 0 ^{- 12} ) ( 0.02 )}{( 8.85 \times 1 0 ^{- 4} )}

C_{B} = 1.8 \times 1 0^{- 9} F

These two capacitors are in parallel. Therefore, net
capacitance of the system is

C = C_{A}^{'} + C_{B} = (0.4 + 1.8) \times 1 0^{- 9} F

C = 2.2 \times 1 0^{- 9} F

Charge stored in the system is

q = q_{A} = 2.2 \times 1 0^{- 7} C

Therefore, energy stored,

U = \frac{1}{2} \frac{q ^{2}}{C}

U = \frac{1}{2} \frac{( 2.2 \times 1 0 ^{- 7} ) ^{2}}{( 2.2 \times 1 0 ^{- 9} )} or U = 1.1 \times 1 0^{- 5} J