MHT CET · Physics · Electrostatics
Two spherical conductors of radii \(4 \mathrm{~cm}\) and \(5 \mathrm{~cm}\) are charged to the same potential. If ' \(\sigma_1\) ' and ' \(\sigma_2\) ' be the respective value of the surface density of charge on the two conductors then the ratio \(\sigma_1: \sigma_2\) is
- A 5:4
- B 3:2
- C 4:3
- D 2:1
Answer & Solution
Correct Answer
(A) 5:4
Step-by-step Solution
Detailed explanation
Spheres are at the same potential
\(
\begin{aligned}
& \therefore \mathrm{V}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}_1}{\mathrm{r}_1}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}_2}{\mathrm{r}_2} \\
& \therefore \frac{\mathrm{q}_1}{\mathrm{q}_2}=\frac{\mathrm{r}_1}{\mathrm{r}_2} \\
& \sigma_1=\frac{\mathrm{q}_1}{4 \pi \mathrm{r}_1^2}, \sigma_2=\frac{\mathrm{q}_2}{4 \pi \mathrm{r}_2^2} \\
& \therefore \frac{\sigma_1}{\sigma_2}=\frac{\mathrm{q}_1}{\mathrm{q}_2} \cdot \frac{\mathrm{r}_2^2}{\mathrm{r}_1^2}=\frac{\mathrm{r}_1}{\mathrm{r}_2} \cdot \frac{\mathrm{r}_2^2}{\mathrm{r}_1^2}=\frac{\mathrm{r}_2}{\mathrm{r}_1}=\frac{5}{4}
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \mathrm{V}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}_1}{\mathrm{r}_1}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}_2}{\mathrm{r}_2} \\
& \therefore \frac{\mathrm{q}_1}{\mathrm{q}_2}=\frac{\mathrm{r}_1}{\mathrm{r}_2} \\
& \sigma_1=\frac{\mathrm{q}_1}{4 \pi \mathrm{r}_1^2}, \sigma_2=\frac{\mathrm{q}_2}{4 \pi \mathrm{r}_2^2} \\
& \therefore \frac{\sigma_1}{\sigma_2}=\frac{\mathrm{q}_1}{\mathrm{q}_2} \cdot \frac{\mathrm{r}_2^2}{\mathrm{r}_1^2}=\frac{\mathrm{r}_1}{\mathrm{r}_2} \cdot \frac{\mathrm{r}_2^2}{\mathrm{r}_1^2}=\frac{\mathrm{r}_2}{\mathrm{r}_1}=\frac{5}{4}
\end{aligned}
\)
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