Chemistry, asked by rajawatsaurabh41, 7 days ago

explain energy loss in the redistribution of charges

Answers

Answered by Anonymous

When two charged conductors joined together through a conducting wire, charge begins to flow from one conductor to another from higher potential to lower potential. ... Due to flow of charge, loss of energy also takes place in the form of heat through the connecting wire.

Answered by ravilaccs

Answer:

When two charged conductors joined together through a conducting wire, charge begins to flow from one conductor to another from higher potential to lower potential. This flow of charge stops when they attain the same potential. Due to flow of charge, loss of energy also takes place in the form of heat through the connecting wire.

Explanation:

Redistribution of Charges and Loss of Energy by Combination of Charged Conductors Two charged conductors which have different potential are combined and then charges on them get redistributed and the potential become same on both of it. In this process, a definite energy is lost. Redistribution of Charges From figure, two isolated conductors A and B are of radii R1 and R2 respectively and the capacitances are C1 and C2 respectively. On charging A by Q1 and B by Q2 , their potentials become V1 and V2 respectively. Before joining the conducting wire, $Q1 = C1V1--- (1) and \\Q2 = C2V2---(2)$ Thus, total amount of charge on both the conductors, $Q = Q1 + Q2 \\Q = C1V1 + C2V2---- (3)$

Now, if both the conductors are attached by a conducting wire of negligible resistance or capacitance. Then charge always flow the conductor of high potential to low potential and flows till the potential on both the conductors is same. In electrical equilibrium, let the potential be V and charges on the conductors be Q’1 and Q’2 respectively. Then, total capacitance on arrangement of the two conductors as a system, C = C1 + C2 because the potential on both the conductors is same and hence the conductors are in parallel combination. After redistribution of charges, Q’1 = C1V ………………. (4) and Q’2 = C2V ……………. (5) Then, total charge after redistribution, Q = Q’1 + Q’2 = (C1 + C2) V From law of conservation of charge, C1V1 + C2V2 = (C1 + C2)V Thus, common potential, Thus, after redistribution of charges, the ratio of charges of conductors, After redistribution of charges, the ratio of charges of conductors is in the ratio of their capacitances. If V1 > V2 Then, V1 > V > V2 Change in potential of first conductor, Hence, after the redistribution of charges, the ratio of change in potential of conductors is equal to the reciprocal of the ratio of their capacitances. Energy Loss In the redistribution process of charges, the charge flows through the conducting wire. Hence, some amount of energy is changed in the form of heat and some amount of energy is lost. It is due to the resistance of conducting wire. Total potential energy of the system before joining conducting wire, (V1 – V2)2 is always positive. Thus, on combining charged conductors and capacitors, in the redistribution of charges, there is always loss of energy. If V1 = V2 Then, ΔC = 0 i.e., There is no flow of charges when potential of both the conductors is same. Hence, there is no loss of energy. A battery of V volt charges the capacitor to give half of energy QV i.e., 1 2 12QV is stored in the capacitor [Note: Q V 2 QV2]and rest half part 1 2 12 QVs lost in the form.

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