An electric dipole is fixed at the origin of coordinates. Its moment is directed in the positive `x`- direction. A positive charge is moved from the point `(r,0)` to the point `(-r,0)` by an external agent. In this process, the work done by the agent is

Options

A. positive and inversely proportional to r
B. positive and inversely proportional to `r^2`
C. negative and inversely proportional to r
D. negative and inversely proportional to `r^2`

Answer

To solve the problem of finding the work done by an external agent when moving a positive charge from point (r, 0) to point (-r, 0) in the presence of a fixed electric dipole, we can follow these steps:

### Step 1: Understand the Configuration
We have an electric dipole fixed at the origin with its dipole moment \( \vec{P} \) directed along the positive x-direction. A positive charge is moved from point \( A (r, 0) \) to point \( B (-r, 0) \).

### Step 2: Calculate the Electric Potential Energy at Points A and B
The electric potential \( V \) due to a dipole at a point \( \vec{r} \) is given by:
\[ V = \frac{k \vec{P} \cdot \hat{r}}{r^2} \]
where \( k \) is a constant, \( \vec{P} \) is the dipole moment, and \( \hat{r} \) is the unit vector in the direction of \( \vec{r} \).

#### At Point A (r, 0):
- The position vector \( \vec{r_A} = (r, 0) \).
- The unit vector \( \hat{r_A} = \hat{i} \).
- The angle \( \theta_A \) between \( \vec{P} \) and \( \hat{r_A} \) is \( 0^\circ \).

Thus, the potential energy \( U_A \) at point A is:
\[ U_A = k \frac{P \cdot 1}{r^2} = \frac{kP}{r^2} \]

#### At Point B (-r, 0):
- The position vector \( \vec{r_B} = (-r, 0) \).
- The unit vector \( \hat{r_B} = -\hat{i} \).
- The angle \( \theta_B \) between \( \vec{P} \) and \( \hat{r_B} \) is \( 180^\circ \).

Thus, the potential energy \( U_B \) at point B is:
\[ U_B = k \frac{P \cdot (-1)}{r^2} = -\frac{kP}{r^2} \]

### Step 3: Calculate the Work Done by the External Agent
The work done \( W \) by the external agent in moving the charge from point A to point B is equal to the change in potential energy:
\[ W = U_B - U_A \]
Substituting the values we found:
\[ W = \left(-\frac{kP}{r^2}\right) - \left(\frac{kP}{r^2}\right) = -\frac{kP}{r^2} - \frac{kP}{r^2} = -\frac{2kP}{r^2} \]

### Step 4: Conclusion
The work done by the external agent is:
\[ W = -\frac{2kP}{r^2} \]