Question 846: Electrostats & Capacitance

Question

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

Accepted Answer

$$V_1=V_2$$ ( $$\because$$ both the conductors have the same potential as they are connected)
$$\begin{aligned}
& \Rightarrow \frac{q_1}{R}=\frac{q_2}{r} \
& \sigma=\frac{q}{4 \pi R^2}
\end{aligned}$$
$$	herefore \quad$$ The ratio of their charge densities is:
$$\begin{aligned}
& \Rightarrow \frac{\sigma_1}{\sigma_2}=\frac{\frac{q_1}{R^2}}{\frac{q_2}{r^2}} \
& 	herefore \quad \frac{\sigma_1}{\sigma_2}=\frac{r}{R}
\end{aligned}$$

Answer

$$\mathrm{R}: \mathrm{r}^2$$

Answer

$$\mathrm{R}: \mathrm{r}$$

Answer

$$\mathrm{r: R}$$

Answer

$$\mathrm{1: 1}$$

Electrostats & Capacitance

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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