A small uncharged metallic sphere is positioned exactly at a point midway between two equal and opposite point charges. If the sphere is slightly displaced towards the positive charge and released, then

Options

A. it will oscillate about its original position
B. it will move further towards the positive charge
C. its potential energy will decrease and kinetic energy will increase
D. the total energy remains constant but is non-zero

Answer

To solve the problem step by step, we will analyze the situation involving the small uncharged metallic sphere placed between two equal and opposite point charges.

### Step 1: Understanding the Initial Setup
We have two point charges:
- Charge \( +q \) (positive charge)
- Charge \( -q \) (negative charge)

These charges are positioned at equal distances from a small uncharged metallic sphere, which is placed at the midpoint between them.

### Step 2: Polarization of the Sphere
When the uncharged metallic sphere is placed in the electric field created by the two point charges, it becomes polarized. This means that:
- The side of the sphere closer to the positive charge will have a slight negative charge induced on it (due to the attraction of electrons).
- The side of the sphere closer to the negative charge will have a slight positive charge induced on it (due to the repulsion of electrons).

### Step 3: Displacement of the Sphere
Now, if we slightly displace the sphere towards the positive charge and release it, we need to analyze the forces acting on it:
- The force due to the positive charge (\( F_1 \)) will attract the induced negative charge on the sphere.
- The force due to the negative charge (\( F_2 \)) will repel the induced positive charge on the sphere.

### Step 4: Analyzing the Forces
When the sphere is displaced towards the positive charge:
- The distance from the positive charge becomes \( r - x \) (where \( x \) is the displacement).
- The distance from the negative charge becomes \( r + x \).

The magnitudes of the forces can be expressed as:
- \( F_1 = k \cdot \frac{q \cdot (-q_{induced})}{(r - x)^2} \)
- \( F_2 = k \cdot \frac{q \cdot (+q_{induced})}{(r + x)^2} \)

Since \( (r - x) < (r + x) \), it follows that \( F_1 > F_2 \). Therefore, the net force acting on the sphere will be directed towards the positive charge.

### Step 5: Motion of the Sphere
As a result of the net force being directed towards the positive charge:
- The sphere will accelerate towards the positive charge.
- It will not oscillate around the midpoint but will continue to move towards the positive charge.

### Step 6: Energy Considerations
As the sphere moves:
- Its potential energy decreases because it is moving in the direction of the electric field created by the positive charge.
- Its kinetic energy increases as it accelerates towards the positive charge.

### Conclusion
Thus, the correct interpretation of the situation is:
- The sphere will not oscillate about its original position; instead, it will move towards the positive charge.
- The potential energy of the sphere will decrease while its kinetic energy will increase.

### Final Answer
The correct option is that the sphere will move towards the positive charge and will not oscillate.

---