Two point charges of +Q each have been placed at the positions `(-a //2,0, 0) and (a // 2, 0,0)`.The locus of the points where - Q charge can be placed such the that total electrostatic potential energy of the system can become equal to zero, is represented by which of the following equations ?

Options

A. A) `Z^(2)+(Y-a)^(2)=2a`
B. B) `Z^(2)+(Y-a)^(2)=27a^(2)//4`
C. C) `Z^(2)+Y^(2)=15a^(2)//4`
D. D) None

Answer

To solve the problem, we need to find the locus of points where a charge of -Q can be placed such that the total electrostatic potential energy of the system becomes zero. We have two positive charges of +Q each located at (-a/2, 0, 0) and (a/2, 0, 0).

### Step-by-Step Solution:

1. **Identify the Positions of Charges**:
- The two positive charges are located at:
- Charge 1: \( Q_1 = +Q \) at \( (-\frac{a}{2}, 0, 0) \)
- Charge 2: \( Q_2 = +Q \) at \( (\frac{a}{2}, 0, 0) \)
- The negative charge \( Q_3 = -Q \) will be placed at a point \( (0, y, z) \).

2. **Calculate Distances**:
- The distance between the two positive charges is \( a \).
- The distance from the negative charge to each positive charge can be calculated using the distance formula:
- Distance from \( Q_1 \) to \( Q_3 \):
\[ r_1 = \sqrt{\left(0 + \frac{a}{2}\right)^2 + y^2 + z^2} = \sqrt{\frac{a^2}{4} + y^2 + z^2} \]
- Distance from \( Q_2 \) to \( Q_3 \):
\[ r_2 = \sqrt{\left(0 - \frac{a}{2}\right)^2 + y^2 + z^2} = \sqrt{\frac{a^2}{4} + y^2 + z^2} \]

3. **Write the Expression for Total Electrostatic Potential Energy**:
- The potential energy \( U \) of the system is given by the formula:
\[ U = k \left( \frac{Q^2}{a} - \frac{Q \cdot (-Q)}{r_1} - \frac{Q \cdot (-Q)}{r_2} \right) \]
- Substituting the distances:
\[ U = k \left( \frac{Q^2}{a} + \frac{Q^2}{\sqrt{\frac{a^2}{4} + y^2 + z^2}} + \frac{Q^2}{\sqrt{\frac{a^2}{4} + y^2 + z^2}} \right) \]
- Simplifying gives:
\[ U = k \left( \frac{Q^2}{a} + 2 \cdot \frac{Q^2}{\sqrt{\frac{a^2}{4} + y^2 + z^2}} \right) \]

4. **Set Total Potential Energy to Zero**:
- For the total potential energy to be zero:
\[ \frac{Q^2}{a} + 2 \cdot \frac{Q^2}{\sqrt{\frac{a^2}{4} + y^2 + z^2}} = 0 \]
- This implies:
\[ \frac{1}{a} = -\frac{2}{\sqrt{\frac{a^2}{4} + y^2 + z^2}} \]

5. **Cross-Multiply and Square Both Sides**:
- Cross-multiplying gives:
\[ \sqrt{\frac{a^2}{4} + y^2 + z^2} = 2a \]
- Squaring both sides results in:
\[ \frac{a^2}{4} + y^2 + z^2 = 4a^2 \]
- Rearranging gives:
\[ y^2 + z^2 = 4a^2 - \frac{a^2}{4} = \frac{16a^2 - a^2}{4} = \frac{15a^2}{4} \]

6. **Final Equation**:
- The equation representing the locus of points where the charge -Q can be placed is:
\[ y^2 + z^2 = \frac{15a^2}{4} \]

### Conclusion:
The correct option that represents the locus of points is:
**Option C: \( z^2 + y^2 = \frac{15a^2}{4} \)**.