To solve the problem of finding the final charge on the smaller sphere after connecting two metallic spheres with given charges, we can follow these steps:
### Step 1: Identify the given values
- Radius of the smaller sphere, \( r_1 = 1 \, \text{cm} = 0.01 \, \text{m} \)
- Radius of the larger sphere, \( r_2 = 2 \, \text{cm} = 0.02 \, \text{m} \)
- Initial charge on the smaller sphere, \( q_1 = 10^{-2} \, \text{C} \)
- Initial charge on the larger sphere, \( q_2 = 5 \times 10^{-2} \, \text{C} \)
### Step 2: Understand the concept of potential
When two conducting spheres are connected by a wire, they reach the same electric potential. The potential \( V \) of a sphere is given by the formula:
\[
V = \frac{k \cdot q}{r}
\]
where \( k \) is Coulomb's constant, \( q \) is the charge, and \( r \) is the radius of the sphere.
### Step 3: Set up the equation for equal potentials
Since the potentials of both spheres must be equal after connecting them:
\[
V_1 = V_2
\]
This gives us:
\[
\frac{k \cdot q_1'}{r_1} = \frac{k \cdot q_2'}{r_2}
\]
Here, \( q_1' \) and \( q_2' \) are the final charges on the smaller and larger spheres, respectively. The \( k \) cancels out:
\[
\frac{q_1'}{r_1} = \frac{q_2'}{r_2}
\]
### Step 4: Relate the charges
From the above equation, we can express \( q_2' \) in terms of \( q_1' \):
\[
q_2' = \frac{r_2}{r_1} \cdot q_1' = \frac{2}{1} \cdot q_1' = 2q_1'
\]
### Step 5: Apply the conservation of charge
According to the law of conservation of charge:
\[
q_1 + q_2 = q_1' + q_2'
\]
Substituting the known values:
\[
10^{-2} + 5 \times 10^{-2} = q_1' + 2q_1'
\]
This simplifies to:
\[
6 \times 10^{-2} = 3q_1'
\]
### Step 6: Solve for \( q_1' \)
Now, we can solve for \( q_1' \):
\[
q_1' = \frac{6 \times 10^{-2}}{3} = 2 \times 10^{-2} \, \text{C}
\]
### Final Answer
The final charge on the smaller sphere is:
\[
\boxed{2 \times 10^{-2} \, \text{C}}
\]
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