Two conducting spheres A and B of radii 4 cm and 2 cm carry charges of `18 xx 10^(-8)` statcoulomb and `9 xx 10^(-8)` statcoulomb respectively of positive electricity. When they are put in electrostatic contact, then the charge will

Options

A. not flow at all
B. flow from A to B
C. flow from B to A
D. disappear

Answer

To solve the problem of two conducting spheres A and B with given charges and radii, we will follow these steps:

### Step 1: Identify the given values
- Radius of sphere A, \( R_A = 4 \, \text{cm} = 0.04 \, \text{m} \)
- Charge on sphere A, \( Q_A = 18 \times 10^{-8} \, \text{C} \)
- Radius of sphere B, \( R_B = 2 \, \text{cm} = 0.02 \, \text{m} \)
- Charge on sphere B, \( Q_B = 9 \times 10^{-8} \, \text{C} \)

### Step 2: Calculate the potential of each sphere
The potential \( V \) of a charged sphere is given by the formula:
\[ V = \frac{kQ}{R} \]
where \( k \) is Coulomb's constant, approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

#### For sphere A:
\[ V_A = \frac{k Q_A}{R_A} = \frac{9 \times 10^9 \times 18 \times 10^{-8}}{0.04} \]
Calculating this:
\[ V_A = \frac{9 \times 10^9 \times 18 \times 10^{-8}}{0.04} = \frac{162 \times 10^1}{0.04} = 4050 \, \text{V} \]

#### For sphere B:
\[ V_B = \frac{k Q_B}{R_B} = \frac{9 \times 10^9 \times 9 \times 10^{-8}}{0.02} \]
Calculating this:
\[ V_B = \frac{9 \times 10^9 \times 9 \times 10^{-8}}{0.02} = \frac{81 \times 10^1}{0.02} = 4050 \, \text{V} \]

### Step 3: Compare the potentials
Now we compare the potentials:
\[ V_A = 4050 \, \text{V} \quad \text{and} \quad V_B = 4050 \, \text{V} \]

Since \( V_A = V_B \), there is no potential difference between the two spheres.

### Step 4: Conclusion
Since there is no potential difference, no charge will flow between the spheres when they are put in electrostatic contact. Therefore, the answer to the question is that the charge will not flow at all.

### Final Answer:
The charge will not flow at all.

---