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

Capacitor `A` with a dielectric
. Can be regarded as two capacitors in parallel, one having a dielectric and the other having no dielectric stare. Such a capacitor has an area of `A//2`. So the combined capacitace is
`C=C_(1)+C_(2)=((A//2)epsilon_(0))/d+((A//2)epsilon_(0)K)/d=A/2(epsilon_(0))/d(1+K)`
`=(0.04xx8.85xx10^(-12))/(2xx8.85xx10^(-4))(1+9)=xx1-0^(-9)F`
Thus, energy stored is
`1/2CV^(2)=1/2xx2xx10^_(9)xx(110)^(2) =1.21xx10^(-5) J`
Work done in removing the dielectric state =(Energy stored in capacitor without dielectric)-(Energy stored in capacitor with dielectric)
It may be noted that on taking out the dielectric, the charge on the capacitor plate remains the same.
`W=q^(2)/(2C')-(q')/(2C)`
Here,
`C = 2 xx 10^(9) F`,
`C'=(Aepsilon_(0))/d=(0.04xx8.85xx10^-12)/(8.85xx10^(-4))=0.4xx10^(-9)F`
q=CV=2xx10^(-9)xx110=2.2xx10^(-7) C`
`:. W=(2.2xx10^(-7))^(2)/2[1/(0.4xx10^(-9))1/(2xx10^(-9)]]`
`=4.84xx10^(-5)J`
ii. The capacitance of B is `epsilon_(0)KA_(B)//d`. so
`C_(B)=1.8xx18^(9) F`
The charge on `A,q_(A)=2.2xx10^(-7) C,` get distributed into two parts `q_(1)` and `q_(2)`. so.
`q_(1)+q_(2)=2.2xx10^(-7) C`
`q_(1)/C_(A)=q_(2)/C_(B)` or `q_(1)=C_(A)/C_(B)q_(2) =(0.4xx10^(-9))/(1.8xx10^(-9))q_(2)=0.22q_(2)`
or `0.22q_(2)+q_(2)=2.2xx10^(-7)`
or `q_(2)=(2.2)/(1.22)xx10^(-7) =1.8xx10^(-7) C`
and `q_(1)=0.4xx10^(-7) C`
Total energy stored is
`q_(1)^(2)/(2C_(A))+q_(1)^(2)/(2C_(B)) =0.2xx10^(-5)+0.9xx10^(-5)=1.1xx10^(-5) J`
Alternatively, the combined capacitance of the two capacitors can be found. The total charge on the two capacitors is known. The energy can be found using the formula `Q^(2)//2C_(eq)`.