Two particles, each of mass m, are a distance d apart. To bring a third particle, also having mass m, from far away to the point midway between the two particles an external agent does work given by:

Options

A. `4Gm^2//d`
B. `-4Gm^2//d`
C. `4Gm^2//d^2`
D. `-4Gm^2//d^2`

Explanation

To solve the problem, we need to calculate the work done by an external agent to bring a third particle of mass \( m \) from far away to the midpoint between the two existing particles, each of mass \( m \), which are separated by a distance \( d \).

### Step-by-Step Solution:

1. **Identify the Position of the Third Particle**:
The third particle is brought to the midpoint between the two particles. The distance from each of the two particles to the midpoint is \( \frac{d}{2} \).

2. **Calculate the Gravitational Potential Energy**:
The gravitational potential energy \( U \) between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by:
\[ U = -\frac{G m_1 m_2}{r} \]
In our case, we have two interactions to consider:
- The interaction between the third particle and the first particle.
- The interaction between the third particle and the second particle.

3. **Potential Energy Between the Third Particle and Each of the Two Existing Particles**:
- For the first particle:
\[ U_1 = -\frac{G m m}{\frac{d}{2}} = -\frac{2G m^2}{d} \]
- For the second particle:
\[ U_2 = -\frac{G m m}{\frac{d}{2}} = -\frac{2G m^2}{d} \]

4. **Total Potential Energy When the Third Particle is at Midpoint**:
The total potential energy \( U_{total} \) when the third particle is at the midpoint is the sum of the potential energies from both interactions:
\[ U_{total} = U_1 + U_2 = -\frac{2G m^2}{d} - \frac{2G m^2}{d} = -\frac{4G m^2}{d} \]

5. **Work Done by the External Agent**:
Since the third particle is brought from a point where the potential energy is zero (far away) to the midpoint, the work done \( W \) by the external agent is equal to the change in potential energy:
\[ W = U_{total} - U_{initial} \]
Here, \( U_{initial} = 0 \) (at infinity):
\[ W = -\frac{4G m^2}{d} - 0 = -\frac{4G m^2}{d} \]

### Final Answer:
The work done by the external agent to bring the third particle to the midpoint is:
\[ W = -\frac{4G m^2}{d} \]