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(C) When two charged conducting spheres are connected by a copper wire,charge flows until they reach a common potential $V$. Since $V = \frac{kQ}{r}$,

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

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(C) When two charged conducting spheres are connected by a copper wire,charge flows until they reach a common potential $V$. Since $V = \frac{kQ}{r}$,for the same potential,$Q \propto r$. Let the new charges be $Q_1$ and $Q_2$ for radii $r_1 = 20\,cm$ and $r_2 = 15\,cm$. Then $\frac{Q_1}{Q_2} = \frac{r_1}{r_2} = \frac{20}{15} = \frac{4}{3}$. The surface charge density is given by $\sigma = \frac{Q}{4\pi r^2}$. Therefore,$\frac{\sigma_1}{\sigma_2} = \frac{Q_1}{Q_2} 	imes \frac{r_2^2}{r_1^2} = \frac{r_1}{r_2} 	imes \frac{r_2^2}{r_1^2} = \frac{r_2}{r_1}$. Substituting the values,$\frac{\sigma_1}{\sigma_2} = \frac{15}{20} = \frac{3}{4}$. Since $\sigma_1 < \sigma_2$,the surface charge density on the smaller sphere $(15\,cm)$ is greater than that on the larger sphere $(20\,cm)$.

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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