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Step-by-Step Solution
Step 1: Identify the Initial Charges and Radii
Two metallic spheres have radii:
Sphere A: radius = 1 cm, initial charge = $-1 \times 10^{-2}\,\text{C}$
Sphere B: radius = 3 cm, initial charge = $5 \times 10^{-2}\,\text{C}$
The total charge before they are connected is:
$$Q = (-1 \times 10^{-2}) + (5 \times 10^{-2}) = 4 \times 10^{-2}\,\text{C}.$$
Step 2: Understand the Equilibration of Potential
When the two spheres are connected by a conducting wire, charges will flow until both spheres have the same electrostatic potential. Let the final charge on the smaller sphere (radius = 1 cm) be $x$, and hence the final charge on the larger sphere (radius = 3 cm) be $Q - x$.
Step 3: Write Down the Condition for Equal Potentials
The potential $V$ of a charged sphere of radius $r$ carrying charge $q$ is given by
$$V = \frac{k\,q}{r},$$
where $k = \frac{1}{4\pi \varepsilon_0}$ is the Coulomb constant. Setting the potentials equal for both spheres:
$$\frac{k\,x}{1} = \frac{k\,(Q - x)}{3}.$$
Step 4: Simplify the Equality
Cancelling $k$ and rearranging, we get:
$$\frac{x}{1} = \frac{Q - x}{3},$$
$$3x = Q - x,$$
$$4x = Q.$$
Step 5: Solve for the Charge on the Smaller Sphere
Substitute $Q = 4 \times 10^{-2}\,\text{C}$:
$$x = \frac{Q}{4} = \frac{4 \times 10^{-2}}{4} = 1 \times 10^{-2}\,\text{C}.$$
Step 6: Obtain the Final Charge on the Bigger Sphere
The charge on the bigger sphere after connection is:
$$Q' = Q - x = 4 \times 10^{-2} - 1 \times 10^{-2} = 3 \times 10^{-2}\,\text{C}.$$
Therefore, the final charge on the bigger sphere is $3 \times 10^{-2}\,\text{C}.$
