- Home
- Standard 12
- Physics
Consider two conducting spheres of radii ${{\rm{R}}_1}$ and ${{\rm{R}}_2}$ with $\left( {{{\rm{R}}_1} > {{\rm{R}}_2}} \right)$. If the two are at the same potential, the larger sphere has more charge than the smaller sphere. State whether the charge density of the smaller sphere is more or less than that of the larger one.
Solution
Both spheres have same potential,
$\therefore \mathrm{V}_{1}=\mathrm{V}_{2}$
$\therefore \frac{k q_{1}}{\mathrm{R}_{1}}=\frac{k q_{2}}{\mathrm{R}_{2}}$
$\therefore \frac{q_{1} \mathrm{R}_{1}}{4 \pi \mathrm{R}_{1}^{2}}=\frac{q_{2} \mathrm{R}_{2}}{4 \pi \mathrm{R}_{2}^{2}} \quad(\because \text { Dividing by } 4 \pi \text { on both sides) }$
$\sigma_{1} \mathrm{R}_{1}=\sigma_{2} \mathrm{R}_{2} \quad\left[\because \sigma=\frac{q}{4 \pi \mathrm{R}^{2}}\right]$
$\text { Now } \mathrm{R}_{1}>\mathrm{R}_{2},$
Charge on larger sphere is more than charge on smaller sphere,
Now, $\frac{k q_{1}}{\mathrm{R}_{1}}=\frac{k q_{2}}{\mathrm{R}_{2}}$ $\frac{\sigma_{1} \mathrm{~A}_{1}}{4 \pi \epsilon_{0} \mathrm{R}_{1}}=\frac{\sigma_{2} \mathrm{~A}_{2}}{4 \pi \epsilon_{0} \mathrm{R}_{2}}\quad[{l}{[\because q=\sigma \mathrm{A}]}$ $\text { and } k=\frac{1}{4 \pi \epsilon_{0}}]$
$\therefore \frac{\sigma_{1} \times 4 \pi \mathrm{R}_{1}^{2}}{4 \pi \epsilon_{0} \mathrm{R}_{1}}=\frac{\sigma_{2} \times 4 \pi \mathrm{R}_{2}^{2}}{4 \pi \epsilon_{0} \mathrm{R}_{2}}$
$\therefore \frac{\sigma_{1} \mathrm{R}_{1}}{\epsilon_{0}}=\frac{\sigma_{2} \mathrm{R}_{2}}{\epsilon_{0}}$
$\therefore \frac{\sigma_{1} R_{1}}{\epsilon_{0}}=\frac{\sigma_{2} R_{2}}{\epsilon_{0}}$ $\therefore \frac{\sigma_{1}}{\sigma_{2}}=\frac{R_{2}}{R_{1}}$ but $R_{1}>R_{2} \Rightarrow 1>\frac{R_{2}}{R_{1}}$ $\frac{\sigma_{1}}{\sigma_{2}}<1$ $\therefore \sigma_{1}<\sigma_{2}$ $\therefore$ Charge density of smaller sphere is less than that of larger sphere.
$\therefore$ Charge density of smaller sphere is less than that of larger sphere.
