A positive charge q is placed in a spherical cavity made in a positively charged sphere. The centres of sphere and cavity are displaced by a small distance `vec(l)`. Force on charge q is :

Options

A. in the direction parallel to vector `vec(l)`
B. in radial direction
C. in a direction which depends on the magnitude of charge density in sphere
D. direction can not be determined

Answer

To solve the problem, we need to analyze the situation where a positive charge \( q \) is placed in a spherical cavity within a positively charged sphere. The centers of the sphere and the cavity are displaced by a small distance \( \vec{l} \). We need to determine the direction of the force acting on the charge \( q \).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a positively charged sphere with a spherical cavity.
- A positive charge \( q \) is placed inside this cavity.
- The centers of the sphere and the cavity are displaced by a small distance \( \vec{l} \).

2. **Electric Field Inside the Sphere**:
- The electric field inside a uniformly charged sphere (with charge density \( \rho \)) at a distance \( r \) from the center is given by:
\[ E_{\text{sphere}} = \frac{\rho r}{3 \epsilon_0} \]
- Here, \( \epsilon_0 \) is the permittivity of free space.

3. **Electric Field Inside the Cavity**:
- The electric field inside the cavity (which is empty) can be considered as zero since there are no charges present inside the cavity.

4. **Resultant Electric Field**:
- The resultant electric field \( E \) at the position of charge \( q \) is the superposition of the electric field due to the sphere and the cavity:
\[ E_{\text{resultant}} = E_{\text{sphere}} - E_{\text{cavity}} = E_{\text{sphere}} - 0 = E_{\text{sphere}} \]
- Therefore, the electric field at the position of charge \( q \) is:
\[ E = \frac{\rho r}{3 \epsilon_0} \]

5. **Displacement Effect**:
- Since the centers of the sphere and the cavity are displaced by a distance \( \vec{l} \), the position of charge \( q \) will also be affected by this displacement.
- The electric field \( E \) will depend on the distance \( l \) because the effective distance from the center of the sphere to the charge \( q \) changes.

6. **Direction of the Force**:
- The force \( F \) on the charge \( q \) due to the electric field \( E \) is given by:
\[ F = qE \]
- Since the electric field \( E \) is directed radially outward from the center of the positively charged sphere, and considering the displacement \( \vec{l} \), the force on charge \( q \) will be in the direction parallel to the displacement \( \vec{l} \).

7. **Conclusion**:
- The force on the charge \( q \) is in the direction parallel to the vector \( \vec{l} \).

### Final Answer:
The correct option is: **The force on charge \( q \) is in the direction parallel to the vector \( \vec{l} \)**.