Two insulated charged spheres of radii 20,cm | vedclass

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(C) When two charged spheres are connected by a conducting wire,charge flows until their potentials become equal $(V_1 = V_2)$.Since the potential of

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Charge on the $25\,cm$ sphere will be greater than that on the $20\,cm$ sphere.

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(C) When two charged spheres are connected by a conducting wire,charge flows until their potentials become equal $(V_1 = V_2)$.Since the potential of a charged sphere is given by $V = \frac{kQ}{R}$,we have $\frac{kQ_1}{R_1} = \frac{kQ_2}{R_2}$.Given $R_1 = 20\,cm$ and $R_2 = 25\,cm$,we get $\frac{Q_1}{20} = \frac{Q_2}{25}$.This implies $\frac{Q_1}{Q_2} = \frac{20}{25} = \frac{4}{5}$.Since the ratio $\frac{Q_1}{Q_2}  Q_1$.Therefore,the charge on the $25\,cm$ sphere $(Q_2)$ will be greater than the charge on the $20\,cm$ sphere $(Q_1)$.

Two insulated charged spheres of radii $20\,cm$ and $25\,cm$ respectively and having an equal charge $Q$ are connected by a copper wire,then they are separated.

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