A large hollow metallic sphere A is charged positively to a potential of 100 volt and a small sphereB to a potential of50 volt. Now B is placed inside A and they are connected by a wire. In which direction will the charge flow ?

Answer

To solve the problem, we need to analyze the situation step by step, focusing on the potentials of the two spheres and how they interact when connected by a wire.

### Step-by-Step Solution:

1. **Identify the potentials of the spheres**:
- Sphere A (the large hollow metallic sphere) is charged positively to a potential of 100 volts.
- Sphere B (the small sphere) is charged to a potential of 50 volts.

2. **Understand the arrangement**:
- Sphere B is placed inside Sphere A and they are connected by a wire.

3. **Recall the properties of conductors**:
- In a conductor, charges redistribute themselves until the potential is uniform throughout the conductor.
- For a hollow metallic sphere, the electric field inside is zero, and the potential inside is constant and equal to the potential on the surface.

4. **Determine the potential inside Sphere A**:
- Since Sphere A is a hollow metallic sphere and is positively charged to 100 volts, the potential inside Sphere A (including the space where Sphere B is located) is also 100 volts.

5. **Analyze the potential of Sphere B**:
- Sphere B has a potential of 50 volts.

6. **Connecting the spheres with a wire**:
- When the wire connects the two spheres, charge will flow from the sphere with higher potential to the sphere with lower potential until equilibrium is reached.

7. **Compare the potentials**:
- The potential of Sphere A (100 volts) is higher than the potential of Sphere B (50 volts).
- However, since Sphere B is inside Sphere A, we must consider the effective potential at Sphere B when connected to Sphere A.

8. **Calculate the effective potential at Sphere B**:
- When Sphere B is inside Sphere A, the potential at Sphere B effectively becomes the sum of its own potential and the potential from Sphere A. Thus, the effective potential at Sphere B becomes:
\[ V_{B\_effective} = V_{B} + V_{A} = 50 \text{ volts} + 100 \text{ volts} = 150 \text{ volts} \]

9. **Determine the direction of charge flow**:
- Since the effective potential at Sphere B (150 volts) is higher than the potential at Sphere A (100 volts), the charge will flow from Sphere B to Sphere A.

### Conclusion:
The charge will flow from Sphere B to Sphere A.

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