In the circuit shown in , capacitor A has capacitance `C_(1)=2 muF` when filled with a dielectric slab `(k=2)`. Capacitors `B` and `C` are air capacitors and have capacitances `C_(2)=3 muF` and `C_(3)=6 muF`, respectively.
.
A is charged by closing the switch `S_(1)` alone.
i. Calculate the energy supplied by the battery during the process of charging.
Switch `S_(1)`is now opened and is closed.
ii. Calculate the charge on B and the energy stored in the system when an electrical equilibrium is atained.
Now switch `S_(2)` is also opened, and the slab of A is removed. Another dielectric slab of `K=2`, which can just fill the space in B, is inserted into it and then switch `S_2` alone is closed.
iii. Calculate by how many time the electric field in B is increased. Calculate also the loss of energy during the redistribution of charge.

Explanation

When switch `S_(1)` alone is closed, capacitor `A` gets directly connected across the battery. Thus, charge on `A` in steady state is
`q_(0) = C_1V =2xx180 =360mu C`
This whole charge is supplied by the battery at emf `V = 180V`. Therefore, the energy supplied by the battery during the charging of capacitor `A` is
`W_("battery") = q_0V = 0.0648 J`
But the energy stored in capacitor `A` is
`U_1 = (1)/(2)q_0v = 0.0324 J`
The remaining part of the energy supplied by the battery is converted into heat during the flow of current through the connecting wires. After `A` is charged switch `S_1` is opened, which disconnects the battery. When `S_2` is closed, some charge is transferred form capacitor `A` to
capacitors `B` and `C`. Let the charge transferred be `q`. In steady state, charges
on the capacitors will be as shown in fig Applying Kirchhoff's loop law,
(1) `(q)/(C_2) +(q)/(C_3) - ((q_0 - q))/(C ) = 0` or `q= 180muC`
Now energy stored in the system of capacitors is
`U_2 = U_A + I_B + U_C = ((q_0 - q)^2)/(2C_1) +(q^2)/(2C_2) +(q^2)/(2C_3) = 0.0162J`
Electric field B is `E = q//epsilon_0A,` where A is the area of plates of capacitor B. Now after the opening of switch `S_2` the dielectric slab of A is removed, therefore, its capacitance becomes
.
`C'_(1)=1 muF`
Another plate of `K=2` is inserted in capacitor B. Therefore, its capacitance becomes `C'_(2)=KC_(2)`
`=6 mu F`.
Now switch `S_(2)` is closed and charge gets redistributed. charge `q'` be further transferred from A to capacitors B to C. In steady state, charges will be as shown in . Applying Kirchhoff's loop law,
(2) `(q+q')/C'_(2)+(q+q')/C_(3)=(q_(0)-q-q')C'_(1)=0 q'=90 muC`
Final charge on capacitor B is `q+q'=27 muC`. Hence, electric field inside it is
`E'=(q+q')/(epsilonKA)`
Factor, by which the electric field in capacitor B is increased is
`(E')/E=((q_q'))/(qK)=0.75`
During the redistribution of charge, energy is lost in the form of heat froduced in the connecting and is equal to Hence, energy lost is
`(-q')^(2)/(2C'_(1))=(q')^(2)/(2C'_(2))+(q')^(2)/(2C3) =5.4xx10^(-3)J`.