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(D) When two charged conductors are brought into contact,charge flows from the conductor at a higher potential to the one at a lower potential until t

Vedclass · Accepted Answer

If $Q_1R_2 
eq Q_2R_1$,the energy of the system decreases.

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(D) When two charged conductors are brought into contact,charge flows from the conductor at a higher potential to the one at a lower potential until their potentials become equal.Let the initial potentials be $V_1 = \frac{kQ_1}{R_1}$ and $V_2 = \frac{kQ_2}{R_2}$.If $V_1 
eq V_2$ (i.e.,$\frac{Q_1}{R_1} 
eq \frac{Q_2}{R_2}$ or $Q_1R_2 
eq Q_2R_1$),charge redistribution occurs.The total energy of the system is $U = \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2$.When connected,the final potential $V$ is given by $V = \frac{Q_1 + Q_2}{C_1 + C_2}$.The final energy is $U' = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2} \frac{(Q_1 + Q_2)^2}{C_1 + C_2}$.It can be mathematically shown that the loss in energy $\Delta U = U - U' = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2$.Since $(V_1 - V_2)^2$ is always positive,$\Delta U > 0$,meaning energy is always lost (decreases) when charge flows between conductors at different potentials.

Two conducting spheres of radii $R_1$ and $R_2$ are charged with charges $Q_1$ and $Q_2$ respectively. When they are brought into contact with each other:

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