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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 capacitaces 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 equilinrium 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. |
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Answer» Solution :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 suppliedby 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 disconnectsthebattery. 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`. 
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