Two spheres of radii 1 cm and 2 cm have been charged with `1.5 xx 10^(-8) and 0.3 xx 10^(-7)` coulomb of posiitve charge. When they are connected with a wire, charge :

Options

A. will flow from the first to the second
B. will flow from the second to the first
C. will not flow at all
D. may flow either from first to second, or from the seoond to first, depending upon the length of the connecting WIRE

Answer

To solve the problem of two charged spheres connected by a wire, we need to determine whether charge will flow between them and, if so, how much. Let's break down the solution step by step.

### Step 1: Identify the Given Data
- **Sphere 1**:
- Radius \( r_1 = 1 \, \text{cm} = 0.01 \, \text{m} \)
- Charge \( Q_1 = 1.5 \times 10^{-8} \, \text{C} \)

- **Sphere 2**:
- Radius \( r_2 = 2 \, \text{cm} = 0.02 \, \text{m} \)
- Charge \( Q_2 = 0.3 \times 10^{-7} \, \text{C} = 3.0 \times 10^{-8} \, \text{C} \)

### Step 2: Calculate the Potentials of Both Spheres
The potential \( V \) at the surface of a charged sphere is given by the formula:
\[ V = \frac{kQ}{R} \]
where \( k \) is Coulomb's constant (\( k \approx 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)), \( Q \) is the charge, and \( R \) is the radius of the sphere.

- **Potential of Sphere 1**:
\[ V_1 = \frac{kQ_1}{r_1} = \frac{k \cdot 1.5 \times 10^{-8}}{0.01} \]

- **Potential of Sphere 2**:
\[ V_2 = \frac{kQ_2}{r_2} = \frac{k \cdot 3.0 \times 10^{-8}}{0.02} \]

### Step 3: Set the Potentials Equal
When the spheres are connected by a wire, the potentials must equalize:
\[ V_1 = V_2 \]

Substituting the expressions for \( V_1 \) and \( V_2 \):
\[ \frac{k \cdot 1.5 \times 10^{-8}}{0.01} = \frac{k \cdot 3.0 \times 10^{-8}}{0.02} \]

### Step 4: Simplify the Equation
Since \( k \) appears on both sides, we can cancel it out:
\[ \frac{1.5 \times 10^{-8}}{0.01} = \frac{3.0 \times 10^{-8}}{0.02} \]

Cross-multiplying gives:
\[ 1.5 \times 10^{-8} \cdot 0.02 = 3.0 \times 10^{-8} \cdot 0.01 \]

Calculating both sides:
- Left Side: \( 1.5 \times 0.02 = 3.0 \times 10^{-10} \)
- Right Side: \( 3.0 \times 0.01 = 3.0 \times 10^{-10} \)

### Step 5: Conclusion
Since both sides are equal, this means that the potentials of both spheres are already equal before they are connected. Therefore, no charge will flow between the spheres when they are connected by a wire.

### Final Answer
No charge will flow between the two spheres when connected by a wire.

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