Two particles A and B having equal charges are placed at distance d apart. A third charged particle placed on the perpendicular bisector at a distance x will experience the maximum Coulomb’s force when :

Options

A. `x = d//sqrt2`
B. `x = d//2`
C. `x = d/(2sqrt2)`
D. `x=d//3sqrt2`

Answer

To solve the problem, we need to find the position of a third charged particle placed on the perpendicular bisector of two equal charges such that it experiences the maximum Coulomb force.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two particles A and B, both with charge \( +q \), placed at a distance \( d \) apart.
- A third charged particle (let's denote it as C) is placed on the perpendicular bisector of the line joining A and B, at a distance \( x \) from the midpoint.

2. **Force Calculation**:
- The force experienced by particle C due to each of the charges A and B can be calculated using Coulomb's law:
\[ F = k \frac{q_1 q_2}{r^2} \]
- Here, \( r \) is the distance from C to either A or B. Since C is on the perpendicular bisector, the distance to both A and B is the same.

3. **Finding the Distance**:
- The distance \( r \) from C to either A or B can be expressed using the Pythagorean theorem:
\[ r = \sqrt{\left(\frac{d}{2}\right)^2 + x^2} \]

4. **Components of the Force**:
- The forces exerted by A and B on C will have components in both x and y directions. However, the x-components will cancel out due to symmetry, and only the y-components will add up.
- The y-component of the force from each charge can be expressed as:
\[ F_y = F \sin(\theta) \]
- Where \( \theta \) is the angle between the line connecting C to A (or B) and the vertical line.

5. **Expressing the Net Force**:
- The net force \( F_{net} \) acting on C due to both charges will be:
\[ F_{net} = 2 F_y = 2 \cdot k \frac{q^2}{r^2} \sin(\theta) \]

6. **Maximizing the Force**:
- To find the maximum force, we need to maximize \( F_{net} \). This involves differentiating \( F_{net} \) with respect to \( x \) and setting the derivative to zero.
- We can express \( \sin(\theta) \) in terms of \( x \) and \( r \):
\[ \sin(\theta) = \frac{x}{r} \]
- Substitute \( r \) into the expression for \( F_{net} \) and differentiate.

7. **Using Trigonometric Identities**:
- We can relate \( \sin(\theta) \) and \( \cos(\theta) \) using the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \).

8. **Finding Critical Points**:
- Set the derivative of \( F_{net} \) with respect to \( x \) to zero to find critical points. This will yield:
\[ \cos(\theta) - 3 \sin^2(\theta) \cos(\theta) = 0 \]
- From this, we can derive \( \sin(\theta) \) and \( \cos(\theta) \).

9. **Final Calculation**:
- After solving, we find that:
\[ \sin(\theta) = \frac{1}{\sqrt{3}} \]
- Using the relationship \( \tan(\theta) = \frac{x}{d/2} \), we can find \( x \):
\[ x = \tan(\theta) \cdot \frac{d}{2} = \frac{1/\sqrt{3}}{\sqrt{2}/\sqrt{2}} \cdot \frac{d}{2} \]
- Simplifying gives us:
\[ x = \frac{d}{2\sqrt{2}} \]

### Conclusion:
The distance \( x \) at which the third charged particle will experience the maximum Coulomb force is:
\[ \boxed{\frac{d}{2\sqrt{2}}} \]