Question

Two charged spherical conductors of radii R₁ and R₂ are connected by a wire. The ratio of surface charge densities of the sphere $\frac{{\mathrm{\sigma }}_1}{{\mathrm{\sigma }}_2}$ will be

• A. $\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}$
• B. $\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}$
• C. $\sqrt{\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}}$
• D. $\frac{{{\mathrm{R}}_1}^2}{{{\mathrm{R}}_2}^2}$

Answer 1

The correct option is B

$\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}$

Step 1: Given data

Radius are ${\mathrm{R}}_1,{\mathrm{R}}_2$.

Step 2: To find

The ratio of surface charge densities

Step 3: Potential will be equal because they are connected to the wire(in series)

$$\begin{gathered} {\mathrm{V}}_1={\mathrm{V}}_2 \\ \frac{{\mathrm{kq}}_1}{{\mathrm{R}}_1}=\frac{{\mathrm{kq}}_2}{{\mathrm{R}}_2} \\ \frac{{\mathrm{q}}_1}{{\mathrm{q}}_2}=\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2} \end{gathered}$$

Here $k$ is constant,$R_1,R_2$ are radius,$q_1,q_2$ are charges, and $V_1,V_2$ are potential.

Step 4: Calculate the ratio of surface charge densities

$$\begin{gathered} \frac{{\mathrm{\sigma }}_1}{{\mathrm{\sigma }}_2}=\frac{{\displaystyle \frac{{\mathrm{q}}_1}{4\mathrm{\pi }{{\mathrm{R}}_1}^2}}}{{\displaystyle \frac{{\mathrm{q}}_2}{4\mathrm{\pi }{{\mathrm{R}}_2}^2}}} \\ =\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}\times \frac{{{\mathrm{R}}_2}^2}{{{\mathrm{R}}_1}^2} \\ =\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1} \end{gathered}$$

Hence, option(B) is correct.