An electric dipole consists of charges `+ 2e and -2e` separated by 0.78mm. It is an electric field of strength `3.4xx10^(6) N//C`. Calculate the magnitude of the torque on the dipole when the dipole moment is (a) parallel to (b) perpendicular to, and (c) antiparallel to the electric field.

Explanation

To solve the problem, we need to calculate the torque on an electric dipole in an electric field for three different orientations: parallel, perpendicular, and antiparallel. The torque (\(\tau\)) on a dipole in an electric field is given by the formula:

\[ \tau = \mathbf{p} \times \mathbf{E} = pE \sin \theta \]

where:
- \( \tau \) is the torque,
- \( \mathbf{p} \) is the dipole moment,
- \( \mathbf{E} \) is the electric field strength,
- \( \theta \) is the angle between the dipole moment and the electric field.

### Step 1: Calculate the Dipole Moment (\( p \))

The dipole moment (\( p \)) is calculated using the formula:

\[ p = Q \cdot d \]

where:
- \( Q \) is the magnitude of one of the charges,
- \( d \) is the separation between the charges.

Given:
- \( Q = 2e = 2 \times 1.6 \times 10^{-19} \, \text{C} = 3.2 \times 10^{-19} \, \text{C} \)
- \( d = 0.78 \, \text{mm} = 0.78 \times 10^{-3} \, \text{m} \)

Calculating \( p \):

\[ p = (3.2 \times 10^{-19} \, \text{C}) \times (0.78 \times 10^{-3} \, \text{m}) = 2.496 \times 10^{-22} \, \text{C m} \]

### Step 2: Calculate Torque for Each Case

#### Case (a): Dipole Moment Parallel to Electric Field (\( \theta = 0^\circ \))

\[ \tau = pE \sin(0^\circ) = pE \cdot 0 = 0 \]

**Torque when parallel = 0 Nm**

#### Case (b): Dipole Moment Perpendicular to Electric Field (\( \theta = 90^\circ \))

\[ \tau = pE \sin(90^\circ) = pE \cdot 1 = pE \]

Given \( E = 3.4 \times 10^6 \, \text{N/C} \):

\[ \tau = (2.496 \times 10^{-22} \, \text{C m}) \times (3.4 \times 10^6 \, \text{N/C}) = 8.49 \times 10^{-16} \, \text{Nm} \]

**Torque when perpendicular = \( 8.49 \times 10^{-16} \, \text{Nm} \)**

#### Case (c): Dipole Moment Antiparallel to Electric Field (\( \theta = 180^\circ \))

\[ \tau = pE \sin(180^\circ) = pE \cdot 0 = 0 \]

**Torque when antiparallel = 0 Nm**

### Final Results

- Torque when parallel to electric field: **0 Nm**
- Torque when perpendicular to electric field: **\( 8.49 \times 10^{-16} \, \text{Nm} \)**
- Torque when antiparallel to electric field: **0 Nm**