An electric dipole is placed at an angle of `30^(@)` with an electric field intensity `2xx10^(5)N//C`. It experiences a torque equal to `4Nm`. The charge on the dipole, if the dipole is length is `2 cm`, is

Options

A. 8 mC
B. 4 mC
C. 6 mC
D. 2 mC

Explanation

To solve the problem, we need to find the charge on the dipole given the torque experienced by it, the electric field intensity, and the angle at which it is placed.

### Step-by-Step Solution:

1. **Understanding the Torque on an Electric Dipole**:
The torque (\( \tau \)) experienced by an electric dipole in an electric field is given by the formula:
\[ \tau = p \cdot E \cdot \sin(\theta) \]
where:
- \( p \) is the dipole moment,
- \( E \) is the electric field intensity,
- \( \theta \) is the angle between the dipole moment and the electric field.

2. **Finding the Dipole Moment**:
The dipole moment (\( p \)) can be expressed in terms of charge (\( q \)) and the separation distance (\( d \)):
\[ p = q \cdot d \]
In this case, the length of the dipole is given as \( 2 \, \text{cm} \), which is \( 2 \times 10^{-2} \, \text{m} \).

3. **Substituting Values into the Torque Equation**:
We know:
- \( \tau = 4 \, \text{Nm} \)
- \( E = 2 \times 10^5 \, \text{N/C} \)
- \( \theta = 30^\circ \) (for which \( \sin(30^\circ) = \frac{1}{2} \))

Substituting these values into the torque formula:
\[ 4 = (q \cdot (2 \times 10^{-2})) \cdot (2 \times 10^5) \cdot \sin(30^\circ) \]
This simplifies to:
\[ 4 = (q \cdot (2 \times 10^{-2})) \cdot (2 \times 10^5) \cdot \frac{1}{2} \]

4. **Simplifying the Equation**:
The equation simplifies to:
\[ 4 = q \cdot (2 \times 10^{-2}) \cdot (10^5) \]
\[ 4 = q \cdot (2 \times 10^{3}) \]

5. **Solving for Charge \( q \)**:
Now, we can solve for \( q \):
\[ q = \frac{4}{2 \times 10^{3}} = \frac{4}{2000} = 0.002 \, \text{C} \]
Converting this to milliCoulombs:
\[ q = 2 \, \text{mC} \]

### Final Answer:
The charge on the dipole is \( 2 \, \text{mC} \).