An electric dipole consists of two opposite charges of magnitude `1 muC` (micro-coulomb) separated by a distance of 2 cm . The dipole is placed in an electric field of `10^(5) V m^(-1)`. (a) What maximum torque does the field exert on the dipole ? (b) How much work must an external agent do to turn the dipole end for end , starting from a position of alignment `theta = 0` ?

Answer

To solve the problem, we will break it down into two parts as specified in the question.

### Part (a): Maximum Torque on the Dipole

1. **Identify the dipole moment (p)**:
The dipole moment \( p \) is given by the formula:
\[ p = q \cdot d \]
where \( q \) is the charge and \( d \) is the separation distance between the charges.

Given:
- \( q = 1 \mu C = 1 \times 10^{-6} C \)
- \( d = 2 cm = 0.02 m \)

Now, calculate \( p \):
\[ p = (1 \times 10^{-6} C) \cdot (0.02 m) = 2 \times 10^{-8} C \cdot m \]

2. **Calculate the maximum torque (\( \tau \))**:
The torque (\( \tau \)) experienced by a dipole in an electric field (\( E \)) is given by:
\[ \tau = p \cdot E \cdot \sin(\theta) \]
The maximum torque occurs when \( \sin(\theta) = 1 \) (i.e., \( \theta = 90^\circ \)):
\[ \tau_{max} = p \cdot E \]

Given:
- \( E = 10^5 \, V/m \)

Now, substitute the values:
\[ \tau_{max} = (2 \times 10^{-8} C \cdot m) \cdot (10^5 \, V/m) = 2 \times 10^{-3} \, N \cdot m \]

### Part (b): Work Done to Turn the Dipole End for End

1. **Use the work done formula**:
The work done \( W \) to rotate the dipole from an angle \( \theta_1 \) to \( \theta_2 \) is given by:
\[ W = p \cdot E \cdot (1 - \cos(\theta)) \]
Here, we are turning the dipole from \( \theta = 0^\circ \) to \( \theta = 180^\circ \).

Thus, we need to calculate \( W \) when \( \theta = 180^\circ \):
\[ W = p \cdot E \cdot (1 - \cos(180^\circ)) \]

Knowing that \( \cos(180^\circ) = -1 \):
\[ W = p \cdot E \cdot (1 - (-1)) = p \cdot E \cdot 2 \]

2. **Substituting the values**:
\[ W = (2 \times 10^{-8} C \cdot m) \cdot (10^5 \, V/m) \cdot 2 \]
\[ W = (2 \times 10^{-3} \, N \cdot m) \cdot 2 = 4 \times 10^{-3} \, J \]

### Final Answers:
(a) Maximum Torque: \( 2 \times 10^{-3} \, N \cdot m \)

(b) Work Done: \( 4 \times 10^{-3} \, J \)