MHT CET · Physics · Current Electricity
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
- A \(\mathrm{R}: \mathrm{r}^2\)
- B \(\mathrm{R}: \mathrm{r}\)
- C \(r: R\)
- D \(1: 1\)
Answer & Solution
Correct Answer
(C) \(r: R\)
Step-by-step Solution
Detailed explanation
\(
V_1=V_2
\)
( \(\because\) 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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