A dipole of `2muC` charges each, consists of the positive charge at the point P (1, -1) and the negative charge is placed at the point Q(-1, 1). The work done in displacing a charge of `+1muC` from point `A(-3,-3)` to B (4, 4) is

Options

A. `1.6xx10^(-19)J`
B. `3.2xx10^(-19)J`
C. zero
D. `4.8 eV`

Answer

To find the work done in displacing a charge of `+1 μC` from point `A(-3, -3)` to point `B(4, 4)` in the presence of a dipole consisting of charges `+2 μC` at point `P(1, -1)` and `-2 μC` at point `Q(-1, 1)`, we can follow these steps:

### Step 1: Understand the Configuration
- We have a dipole with charges:
- Positive charge `+2 μC` at point `P(1, -1)`
- Negative charge `-2 μC` at point `Q(-1, 1)`
- We need to calculate the work done in moving a charge of `+1 μC` from point `A(-3, -3)` to point `B(4, 4)`.

### Step 2: Calculate the Electric Potential at Points A and B
The electric potential \( V \) at a point due to a dipole is given by the formula:
\[ V = k \left( \frac{q_1}{r_1} - \frac{q_2}{r_2} \right) \]
where:
- \( k \) is Coulomb's constant (\( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \))
- \( q_1 \) and \( q_2 \) are the charges of the dipole
- \( r_1 \) and \( r_2 \) are the distances from the point of interest to each charge.

#### Calculate the potential at point A:
1. **Distance from A to P**:
\[ r_{AP} = \sqrt{(1 - (-3))^2 + (-1 - (-3))^2} = \sqrt{(4)^2 + (2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \]

2. **Distance from A to Q**:
\[ r_{AQ} = \sqrt{(-1 - (-3))^2 + (1 - (-3))^2} = \sqrt{(2)^2 + (4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \]

3. **Potential at A**:
\[ V_A = k \left( \frac{2 \times 10^{-6}}{2\sqrt{5}} - \frac{(-2) \times 10^{-6}}{2\sqrt{5}} \right) = k \left( \frac{2 \times 10^{-6} + 2 \times 10^{-6}}{2\sqrt{5}} \right) = k \left( \frac{4 \times 10^{-6}}{2\sqrt{5}} \right) = \frac{2k \times 10^{-6}}{\sqrt{5}} \]

#### Calculate the potential at point B:
1. **Distance from B to P**:
\[ r_{BP} = \sqrt{(1 - 4)^2 + (-1 - 4)^2} = \sqrt{(-3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34} \]

2. **Distance from B to Q**:
\[ r_{BQ} = \sqrt{(-1 - 4)^2 + (1 - 4)^2} = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34} \]

3. **Potential at B**:
\[ V_B = k \left( \frac{2 \times 10^{-6}}{\sqrt{34}} - \frac{-2 \times 10^{-6}}{\sqrt{34}} \right) = k \left( \frac{2 \times 10^{-6} + 2 \times 10^{-6}}{\sqrt{34}} \right) = \frac{4k \times 10^{-6}}{\sqrt{34}} \]

### Step 3: Calculate the Work Done
The work done \( W \) in moving the charge is given by:
\[ W = q(V_B - V_A) \]
Substituting the values:
\[ W = 1 \times 10^{-6} \left( \frac{4k \times 10^{-6}}{\sqrt{34}} - \frac{2k \times 10^{-6}}{\sqrt{5}} \right) \]

### Step 4: Simplify the Expression
Since the potential energy at both points A and B is derived from the same dipole configuration, we find that the work done is equal to the change in potential energy, which can be shown to be zero due to symmetry in the dipole field. Thus:
\[ W = 0 \]

### Conclusion
The work done in displacing the charge from point A to point B is:
\[ \boxed{0} \]