The charges - q, Q, -q are placed along x-axis at x = 0, x = a and x = 2a. If the potential energy of the system is zero, then Q:q is

Options

A. `1 : 2`
B. `2 : 1`
C. `1 : 4`
D. `4 : 1`

Answer

To solve the problem, we need to find the ratio \( \frac{Q}{q} \) such that the potential energy of the system of charges is zero. We have three charges: \( -q \) at \( x = 0 \), \( Q \) at \( x = a \), and \( -q \) at \( x = 2a \).

### Step-by-Step Solution:

1. **Identify the positions and charges**:
- Charge \( -q \) is at \( x = 0 \)
- Charge \( Q \) is at \( x = a \)
- Charge \( -q \) is at \( x = 2a \)

2. **Calculate the potential energy \( U \) of the system**:
The potential energy \( U \) of a system of point charges is given by the formula:
\[ U = k \sum_{i < j} \frac{q_i q_j}{r_{ij}} \]
where \( k = \frac{1}{4\pi\epsilon_0} \).

3. **Calculate the potential energy contributions**:
- The potential energy between \( -q \) at \( x = 0 \) and \( Q \) at \( x = a \):
\[ U_{1} = \frac{-qQ}{a} \]
- The potential energy between \( Q \) at \( x = a \) and \( -q \) at \( x = 2a \):
\[ U_{2} = \frac{-qQ}{a} \]
- The potential energy between the two \( -q \) charges at \( x = 0 \) and \( x = 2a \):
\[ U_{3} = \frac{(-q)(-q)}{2a} = \frac{q^2}{2a} \]

4. **Combine the potential energy contributions**:
The total potential energy \( U \) of the system is:
\[ U = U_{1} + U_{2} + U_{3} = \frac{-qQ}{a} + \frac{-qQ}{a} + \frac{q^2}{2a} \]
Simplifying this gives:
\[ U = \frac{-2qQ}{a} + \frac{q^2}{2a} \]

5. **Set the total potential energy to zero**:
According to the problem, the potential energy \( U \) is zero:
\[ \frac{-2qQ}{a} + \frac{q^2}{2a} = 0 \]

6. **Multiply through by \( 2a \) to eliminate the denominator**:
\[ -4qQ + q^2 = 0 \]

7. **Rearranging gives**:
\[ q^2 = 4qQ \]

8. **Divide both sides by \( q \) (assuming \( q \neq 0 \))**:
\[ q = 4Q \]

9. **Express \( \frac{Q}{q} \)**:
\[ \frac{Q}{q} = \frac{1}{4} \]

### Final Answer:
Thus, the ratio \( \frac{Q}{q} \) is \( \frac{1}{4} \).