Two equal point charges are fixed at `x=-a` and `x=+a` on the x-axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to

Options

A. (a) x
B. (b) `x^2`
C. (c) `x^3`
D. (d) `1/x`

Explanation

To solve the problem, we need to determine the change in electrical potential energy of a point charge \( Q \) when it is displaced a small distance \( x \) along the x-axis, given that two equal point charges are fixed at \( x = -a \) and \( x = +a \).

### Step-by-step Solution:

1. **Identify the Initial Potential Energy**:
The initial potential energy \( U_i \) of the charge \( Q \) at the origin (0, 0) due to the two fixed charges \( q \) at \( x = -a \) and \( x = +a \) can be calculated using the formula for the potential energy between two point charges:
\[ U_i = k \left( \frac{qQ}{a} + \frac{qQ}{a} \right) = 2 \frac{kqQ}{a} \]
where \( k = \frac{1}{4\pi\epsilon_0} \).

2. **Determine the Final Potential Energy**:
When the charge \( Q \) is displaced by a small distance \( x \), the distances to the two charges change. The new distances are \( a - x \) and \( a + x \). Thus, the final potential energy \( U_f \) is:
\[ U_f = k \left( \frac{qQ}{a - x} + \frac{qQ}{a + x} \right) \]

3. **Calculate the Change in Potential Energy**:
The change in potential energy \( \Delta U \) is given by:
\[ \Delta U = U_f - U_i \]
Substituting the expressions for \( U_f \) and \( U_i \):
\[ \Delta U = k \left( \frac{qQ}{a - x} + \frac{qQ}{a + x} \right) - 2 \frac{kqQ}{a} \]

4. **Simplifying the Expression**:
To simplify \( \Delta U \), we can find a common denominator for the terms in \( U_f \):
\[ \Delta U = kqQ \left( \frac{(a + x) + (a - x)}{(a - x)(a + x)} - \frac{2a}{a} \right) \]
This simplifies to:
\[ \Delta U = kqQ \left( \frac{2a}{a^2 - x^2} - 2 \right) \]

5. **Further Simplification Using Approximation**:
Since \( x \) is small compared to \( a \) (i.e., \( x \ll a \)), we can approximate \( a^2 - x^2 \approx a^2 \). Thus:
\[ \Delta U \approx kqQ \left( \frac{2a}{a^2} - 2 \right) = kqQ \left( \frac{2}{a} - 2 \right) = kqQ \left( \frac{2 - 2a}{a} \right) \]

6. **Final Expression for Change in Potential Energy**:
The dominant term in the change in potential energy as \( x \) becomes small is:
\[ \Delta U \approx \frac{2kqQ}{a} x^2 \]
Therefore, the change in electrical potential energy \( \Delta U \) is approximately proportional to \( x^2 \).

### Conclusion:
The change in electrical potential energy of the charge \( Q \) when displaced by a small distance \( x \) is approximately proportional to \( x^2 \).