Two parallel pltate capacitors A and B have the same separation `d=8.85xx10^(-4)m` between the plates. The plate areas of A and B are `0.04m^(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.
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i. The dielectric slab placed insid A as shown in. A is then charge to a potential difference of `110 V. Calculate the capacitance of A and the energy stored in it.
ii. The battery is disconnected, and then the dielectric slab is removed from A Find the work done by the extergy agency in removing the slab from A.
iii. The same dielectric slab is now placed inside B, filling it completely. The two capcitors `A and B` are then connected as shown in. `(c )`. Calculate the energy stored in the system.

Explanation

(i) capacitor A is a combination of two capacitor `C_K` and `C_0` in parallel.Hence,
`C_A=C_K+C_0=(Kepsilon_0A)/d+(epsilon_0A)/d=(K+1)(epsilon_0A)/d`
Here, `A=0.22 m^(2)` Substituting the values,we have
`C_A=(9+1)(8.85xx10^(-12)(0.02))/((8.85xx10^(-4)))C_A=2.0xx10^(-9)F`
Energy stored in capacitor,A when connected with a 110V battery is
`U _A=1/2C _AV^(2)=1/2(2xx10^(-9))(110)^(2) U _A=1.21xx10^(-5)J`
Charge stored in the capacitor, `q_A=C_AV(2.0xx10^(-9))(110)q_A =2.2xx10^(-7)C`
Now,this charge remains constant even after battry is disconnected but when the slab is removed.
Capacitance of A will get reduced. Let be `C'A`
`C'A=(epsilon_0(2A))/d=((8.85xx10^(-12))/(8.85xx10^(-4)))C' A =0.4xx10^(-9) F` Energy stored in this case would be
`U'_ A=1/2(q_A)^(2)/(C'_A)=1/2((2.2xx10^(-7))^(2)/((0.4xx10^(-9)))) U '_A=6.05xx10^(-5)J lt U _A`
Therefore, work done to remove the slab would be
`W=U '_A-U_A=(6.05xx1.21)xx10^(-5)J or W=4.84xx10^(-5)J`
(iii) Capacity of B when filled with dielectric ,`C_B=(Kepsilon_0A)/(d)=((9)(8.85xx10^(-12))(0.02))/((8.85xx10^(-4)))`
`C_B=1.8xx10^(-9)F`, Thes two capacis are in parallel,therefore,net capacitance of the system is
`C=C' _A+C _B=(0.4+1.8)xx10^(-9)F C=2.2 xx10^(-9)F` , Change stored in the system is `q=q_A-2.2xx10^(-7)C`
Threfore,energy stored,`U=1/2(q^(2))/C, U=1/2((2.2xx10^(-7))^(2))/((2.2xx10^(-9)))or U=1.1xx10^(-5) J`