Two metal spheres of capacitance C_1 and C_2 | vedclass

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

Vedclass · Accepted Answer

$\frac{Q_1}{Q_2} = \frac{C_1}{C_2}$

Vedclass · Answer

(B) When two conductors are brought into contact,charge flows from the conductor at a higher potential to the conductor at a lower potential until their potentials become equal.Let the final common potential be $V$.The charge on the first sphere is $Q_1 = C_1 V$.The charge on the second sphere is $Q_2 = C_2 V$.Taking the ratio of the two charges,we get $\frac{Q_1}{Q_2} = \frac{C_1 V}{C_2 V} = \frac{C_1}{C_2}$.Therefore,the final charges on the spheres satisfy the relation $\frac{Q_1}{Q_2} = \frac{C_1}{C_2}$.

Two metal spheres of capacitance $C_1$ and $C_2$ carry some charges. They are put in contact and then separated. The final charges $Q_1$ and $Q_2$ on them will satisfy

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