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(D) The potential of a conducting sphere is given by $V = \frac{Q}{4 \pi \varepsilon_0 R}$.When two spheres are brought into contact,charge flows unti

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a decrease in energy of the system if $Q_1R_2 
eq Q_2R_1$

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(D) The potential of a conducting sphere is given by $V = \frac{Q}{4 \pi \varepsilon_0 R}$.When two spheres are brought into contact,charge flows until their potentials become equal.The initial energy of the system is $U_i = \frac{1}{2} \frac{Q_1^2}{C_1} + \frac{1}{2} \frac{Q_2^2}{C_2}$,where $C_1 = 4 \pi \varepsilon_0 R_1$ and $C_2 = 4 \pi \varepsilon_0 R_2$.The final energy of the system is $U_f = \frac{1}{2} (C_1 + C_2) V^2$,where $V = \frac{Q_1 + Q_2}{C_1 + C_2}$.The loss in energy is given by $\Delta U = U_i - U_f = \frac{C_1 C_2}{2(C_1 + C_2)} (V_1 - V_2)^2$.Since $(V_1 - V_2)^2$ is always positive,$\Delta U > 0$ if $V_1 
eq V_2$.$V_1 = V_2$ implies $\frac{Q_1}{R_1} = \frac{Q_2}{R_2}$,which means $Q_1 R_2 = Q_2 R_1$.Therefore,if $Q_1 R_2 
eq Q_2 R_1$,there is a decrease in the energy of the system.

Two conducting spheres of radii $R_1$ and $R_2$ are charged with charges $Q_1$ and $Q_2$ respectively. On bringing them in contact,there is:

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