Two insulated charged conducting spheres of r | vedclass

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(c) After redistribution, charges on them will be different, but they will acquire common potential
i.e. $k\frac{{{Q_1}}}{{{r_1}}} = k\frac{{{Q_2}}}{

Vedclass · Accepted Answer

Surface charge density on the $15\,cm$ sphere will be greater than that on the $20\,cm$ sphere

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(c) After redistribution, charges on them will be different, but they will acquire common potential
i.e. $k\frac{{{Q_1}}}{{{r_1}}} = k\frac{{{Q_2}}}{{{r_2}}}$ $==>$ $\frac{{{Q_1}}}{{{Q_2}}} = \frac{{{r_1}}}{{{r_2}}}$
As $\sigma = \frac{Q}{{4\pi \,{r^2}}}$ $==>$ $\frac{{{\sigma _1}}}{{{\sigma _2}}} = \frac{{{Q_1}}}{{{Q_2}}} 	imes \frac{{r_2^2}}{{r_1^2}}$ $==>$ $\frac{{{\sigma _1}}}{{{\sigma _2}}} = \frac{{{r_2}}}{{{r_1}}}$ $==>$ $\sigma \propto \frac{1}{r}$
i.e. surface charge density on smaller sphere will be more.

Two insulated charged conducting spheres of radii $20\,cm$ and $15\,cm$ respectively and having an equal charge of $10\,C$ are connected by a copper wire and then they are separated. Then

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