Question

Two conducting charged spheres having different radii
(Solid or Hollow) are connected by a conducting wire. Then
which of the following is true
А
Finally both bodies will acquire equal amount of
charge
В
Finally bigger sphere has more charge than
smaller sphere
©
Finally bigger sphere has less charge than
smaller sphere
D
Any of the above situation can appear​

Answer 1

Answer:

I think your answer is option C is true

Answer 2

Answer:

The correct answer is option C

Explanation:

• Whenever two conducting charged spheres X &Y having equal charges are connected.
• The charge is always conserved, it can never be created nor destroyed
• V = kQ/r so the smaller sphere is at the lower potential (more negative = lower) Negative charge flows from low to high potential so the charge will flow from the smaller sphere to the larger. The flow of charge ceases when there is no difference in potential.
• Here, $\quad \mathrm{V}_{1}=\mathrm{V}_{2} \quad or \quad \frac{\mathrm{q}_{1}}{4 \pi \varepsilon_{0} \mathrm{R}_{1}}=\frac{\mathrm{q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}_{2}}$
• $\therefore \frac{\mathrm{q}_{1}}{\mathrm{q}_{2}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}} \ldots(\mathrm{i})$
• Given $\mathrm{R}_{1} > \mathrm{R}_{2}$
• $\therefore \mathrm{q}_{1} > \mathrm{q}_{2}$
• $\therefore$Larger sphere has more charge than the smaller sphere. Now charge densities
• $\sigma_{1}=\frac{\mathrm{q}_{1}}{4 \pi \mathrm{R}_{1}^{2}} ;\\ \sigma_{2}=\frac{\mathrm{q}_{2}}{4 \pi \mathrm{R}_{2}^{2}}$
• $\therefore \frac{\sigma_{2}}{\sigma_{1}}=\frac{\mathrm{q}_{2}}{\mathrm{q}_{1}} \frac{\mathrm{R}_{2}^{2}}{\mathrm{R}_{2}^{2}} \quad \text { or } \\\quad \frac{\sigma_{2}}{\sigma_{1}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}} \frac{\mathrm{R}_{1}^{2}}{\mathrm{R}_{2}^{2}}\\=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}(\text { using (i)) }$
• As $\mathrm{R}_{1} > \mathrm{R}_{2} ,$ therefore $\sigma_{2} > \sigma_{2}$
• Charge density of smaller sphere is more than the charge density of larger sphere