Two metallic spheres of radii `1 cm` and `2 cm` are given charges `10^(-2) C` and `5 xx 10^(-2) C` respectively. If they are connected by a conducting wire, the final charge on the smaller sphere is

Options

A. `3xx10^(-2)C`
B. `4xx10^(-2)C`
C. `1xx10^(-2)C`
D. `2xx10^(-2)C`

Explanation

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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