Two parallel plate capacitors A and B have the same separation `d=8.85xx10^-4m` between the plates. The plate area of A and B are `0.04m^2` and `0.02m^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 figure (a). A is then charged to a potential difference of 110V. Calculate the capacitance of A and the energy stored in it.
The battery is disconnected and then the dielectric slab is moved from A. Find the work done by the external agency in removing the slab from A.
(iii) 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(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`