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A charged spherical conductor of radius ' $R$ ' is connected momentarily to another uncharged spherical conductor of radius ' $r$ ' by means of a thin conducting wire, then the ratio of the surface charge density of the first to the second conductor is
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Verified Answer
The correct answer is:
$r: R$
$$
V_1=V_2
$$
( $\because$ both the conductors have the same potential as they are connected)
$$
\begin{gathered}
\Rightarrow \frac{q_1}{R}=\frac{q_2}{r} \\
\sigma=\frac{q}{4 \pi R^2}
\end{gathered}
$$
$\therefore \quad$ The ratio of their charge densities is:
$$
\begin{gathered}
\Rightarrow \frac{\sigma_1}{\sigma_2}=\frac{\frac{\mathrm{q}_1}{\mathrm{R}^2}}{\frac{\mathrm{q}_2}{\mathrm{r}^2}} \\
\therefore \quad \frac{\sigma_1}{\sigma_2}=\frac{\mathrm{r}}{\mathrm{R}}
\end{gathered}
$$
V_1=V_2
$$
( $\because$ both the conductors have the same potential as they are connected)
$$
\begin{gathered}
\Rightarrow \frac{q_1}{R}=\frac{q_2}{r} \\
\sigma=\frac{q}{4 \pi R^2}
\end{gathered}
$$
$\therefore \quad$ The ratio of their charge densities is:
$$
\begin{gathered}
\Rightarrow \frac{\sigma_1}{\sigma_2}=\frac{\frac{\mathrm{q}_1}{\mathrm{R}^2}}{\frac{\mathrm{q}_2}{\mathrm{r}^2}} \\
\therefore \quad \frac{\sigma_1}{\sigma_2}=\frac{\mathrm{r}}{\mathrm{R}}
\end{gathered}
$$
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