Two point charges q each are fixed at (a,0) and (-a,0). A third charge Q is placed at origin. Electrons potential energy of the system will

Options

A. increase if `Q` is slightly displaced along x-axis
B. decrease if Q is slightly displaced along x-axis
C. increase if Q is slightly displaced along y-axis
D. decrease if Q is slightly displaced along y-axis

Answer

To determine the change in the electric potential energy of the system when the charge \( Q \) is displaced, we can follow these steps:

### Step 1: Identify the Configuration
We have two point charges \( q \) fixed at positions \( (a, 0) \) and \( (-a, 0) \), and a third charge \( Q \) placed at the origin \( (0, 0) \).

### Step 2: Calculate Initial Potential Energy
The initial potential energy \( U_{\text{initial}} \) of the system can be calculated by considering the interactions between the charges. The potential energy between two point charges is given by the formula:
\[ U = k \frac{q_1 q_2}{r} \]
where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them.

1. The potential energy between \( Q \) and the charge at \( (a, 0) \):
\[ U_{Qq_1} = k \frac{Qq}{a} \]

2. The potential energy between \( Q \) and the charge at \( (-a, 0) \):
\[ U_{Qq_2} = k \frac{Qq}{a} \]

3. The potential energy between the two charges \( q \) at \( (a, 0) \) and \( (-a, 0) \):
\[ U_{qq} = k \frac{q \cdot q}{2a} = k \frac{q^2}{2a} \]

Thus, the total initial potential energy is:
\[ U_{\text{initial}} = U_{Qq_1} + U_{Qq_2} + U_{qq} = k \frac{Qq}{a} + k \frac{Qq}{a} + k \frac{q^2}{2a} = 2k \frac{Qq}{a} + k \frac{q^2}{2a} \]

### Step 3: Displace Charge \( Q \)
Now, let's analyze the displacement of charge \( Q \).

#### Case 1: Displacement along the x-axis
If \( Q \) is displaced slightly along the x-axis by a distance \( \delta x \):
- The distance to the charge at \( (a, 0) \) becomes \( a - \delta x \).
- The distance to the charge at \( (-a, 0) \) becomes \( a + \delta x \).

The new potential energy \( U_{\text{final}} \) will be:
\[ U_{\text{final}} = k \frac{Qq}{a - \delta x} + k \frac{Qq}{a + \delta x} + k \frac{q^2}{2a} \]

Using the approximation for small \( \delta x \):
\[ \frac{1}{a - \delta x} \approx \frac{1}{a} + \frac{\delta x}{a^2}, \quad \frac{1}{a + \delta x} \approx \frac{1}{a} - \frac{\delta x}{a^2} \]

Thus:
\[ U_{\text{final}} \approx kQq \left( \frac{1}{a} + \frac{\delta x}{a^2} + \frac{1}{a} - \frac{\delta x}{a^2} \right) + k \frac{q^2}{2a} = 2k \frac{Qq}{a} + k \frac{q^2}{2a} \]

This shows that \( U_{\text{final}} > U_{\text{initial}} \) when \( Q \) is displaced along the x-axis, indicating that the potential energy increases.

#### Case 2: Displacement along the y-axis
If \( Q \) is displaced slightly along the y-axis by a distance \( \delta y \), the distances to both charges \( q \) increase:
- The distance to both charges becomes \( \sqrt{a^2 + (\delta y)^2} \).

The new potential energy will be:
\[ U_{\text{final}} = 2k \frac{Qq}{\sqrt{a^2 + (\delta y)^2}} + k \frac{q^2}{2a} \]

For small \( \delta y \):
\[ \sqrt{a^2 + (\delta y)^2} \approx a + \frac{(\delta y)^2}{2a} \]
Thus:
\[ U_{\text{final}} < U_{\text{initial}} \]
indicating that the potential energy decreases when \( Q \) is displaced along the y-axis.

### Conclusion
- The potential energy increases if \( Q \) is displaced along the x-axis.
- The potential energy decreases if \( Q \) is displaced along the y-axis.