Two metallic spheres of radii 1 cm and 3 cm are given charges of -1×10⁻² C and 5×10⁻² C, respectively. If thes

Question

Two metallic spheres of radii 1 cm and 3 cm are given charges of -1×10⁻² C and 5×10⁻² C, respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is:

Accepted Answer

1. **Conservation of Charge:** When the two spheres are connected, the total charge is conserved. 
   $Q_{total} = Q_1 + Q_2 = (-1 	imes 10^{-2}) + (5 	imes 10^{-2}) = 4 	imes 10^{-2} 	ext{ C}$.

2. **Common Potential:** Charge flows until the electric potential is the same on both spheres ($V_1 = V_2$). For a spherical conductor, Potential $V = kQ/R$. Thus, $Q'_1/R_1 = Q'_2/R_2$, which implies that the final charge is proportional to the radius ($Q \propto R$).

3. **Redistribution Formula:** The final charge on a sphere is a fraction of the total charge based on its capacitance (or radius).
   $Q'_{bigger} = Q_{total} 	imes \frac{R_{bigger}}{R_{small} + R_{bigger}}$

4. **Calculation:**
   Given $R_1 = 1 	ext{ cm}$, $R_2 = 3 	ext{ cm}$ (bigger sphere).
   $Q'_{bigger} = (4 	imes 10^{-2}) 	imes \frac{3}{1 + 3}$
   $Q'_{bigger} = (4 	imes 10^{-2}) 	imes \frac{3}{4} = 3 	imes 10^{-2} 	ext{ C}$.

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