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. `3 xx 10^(-2) C`
B. `1 xx 10^(-2) C`
C. `4 xx 10^(-2) C`
D. `2 xx 10^(-2) C`

Explanation

To solve the problem, we need to determine the final charge on the smaller sphere after they are connected by a conducting wire. Here is the step-by-step solution:

### Step 1: Understand the initial conditions
We have two metallic spheres:
- Sphere 1 (smaller): Radius \( r_1 = 1 \, \text{cm} = 0.01 \, \text{m} \) with charge \( Q_1 = 10^{-2} \, \text{C} \)
- Sphere 2 (larger): Radius \( r_2 = 2 \, \text{cm} = 0.02 \, \text{m} \) with charge \( Q_2 = 5 \times 10^{-2} \, \text{C} \)

### Step 2: Calculate the total charge
The total charge \( Q_{\text{total}} \) when the spheres are connected is:
\[ Q_{\text{total}} = Q_1 + Q_2 = 10^{-2} + 5 \times 10^{-2} = 6 \times 10^{-2} \, \text{C} \]

### Step 3: Set up the equation for equal potential
When the spheres are connected, they will share the total charge until their potentials are equal. The potential \( V \) of a sphere is given by the formula:
\[ V = \frac{kQ}{R} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge on the sphere, and \( R \) is the radius of the sphere.

Let \( Q_f \) be the final charge on the smaller sphere. Therefore, the charge on the larger sphere will be \( Q_{\text{total}} - Q_f \).

### Step 4: Write the equation for equal potentials
The potentials of the two spheres must be equal:
\[ \frac{k Q_f}{r_1} = \frac{k (Q_{\text{total}} - Q_f)}{r_2} \]

### Step 5: Simplify the equation
Since \( k \) cancels out, we can simplify the equation:
\[ \frac{Q_f}{r_1} = \frac{Q_{\text{total}} - Q_f}{r_2} \]

Substituting the values of \( r_1 \) and \( r_2 \):
\[ \frac{Q_f}{0.01} = \frac{6 \times 10^{-2} - Q_f}{0.02} \]

### Step 6: Cross-multiply and solve for \( Q_f \)
Cross-multiplying gives:
\[ 0.02 Q_f = 0.01 (6 \times 10^{-2} - Q_f) \]
Expanding this:
\[ 0.02 Q_f = 0.006 - 0.01 Q_f \]
Combining like terms:
\[ 0.02 Q_f + 0.01 Q_f = 0.006 \]
\[ 0.03 Q_f = 0.006 \]
Dividing both sides by \( 0.03 \):
\[ Q_f = \frac{0.006}{0.03} = 0.2 \, \text{C} \]

### Step 7: Final charge on the smaller sphere
Thus, the final charge on the smaller sphere is:
\[ Q_f = 0.2 \, \text{C} \]