An electric dipole of length `2cm` is placed with its axis making an angle `30^(@)` to a uniform electric field `10^(5)N/C` .If it experiences a torque of `10sqrt(3)`Nm ,then potential energy of the dipole ..

Options

A. `- 10 ` J
B. `-20 ` J
C. `-30` J
D. `-40 `J

Answer

To solve the problem step by step, we will follow the concepts of electric dipoles, torque, and potential energy in an electric field.

### Step 1: Understand the given data
- Length of the dipole, \( l = 2 \, \text{cm} = 0.02 \, \text{m} \)
- Angle with the electric field, \( \theta = 30^\circ \)
- Electric field strength, \( E = 10^5 \, \text{N/C} \)
- Torque experienced by the dipole, \( \tau = 10\sqrt{3} \, \text{Nm} \)

### Step 2: Relate torque to dipole moment and electric field
The torque \( \tau \) experienced by a dipole in an electric field is given by the formula:
\[ \tau = p \cdot E \cdot \sin(\theta) \]
where \( p \) is the dipole moment, and \( \theta \) is the angle between the dipole moment and the electric field.

### Step 3: Calculate the dipole moment
The dipole moment \( p \) is defined as:
\[ p = q \cdot l \]
However, we do not need to find \( p \) directly to solve for potential energy.

### Step 4: Relate potential energy to dipole moment and electric field
The potential energy \( U \) of a dipole in an electric field is given by:
\[ U = -p \cdot E \cdot \cos(\theta) \]

### Step 5: Divide the equations for torque and potential energy
From the torque equation:
\[ \tau = p \cdot E \cdot \sin(30^\circ) \]
And for potential energy:
\[ U = -p \cdot E \cdot \cos(30^\circ) \]

Dividing the torque equation by the potential energy equation gives:
\[ \frac{\tau}{U} = -\frac{\sin(30^\circ)}{\cos(30^\circ)} = -\tan(30^\circ) \]

### Step 6: Substitute known values
We know:
- \( \sin(30^\circ) = \frac{1}{2} \)
- \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \)
- \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \)

Thus, we can rewrite:
\[ \tau = 10\sqrt{3} \, \text{Nm} \]
Substituting into the equation:
\[ \frac{10\sqrt{3}}{U} = -\frac{1}{\sqrt{3}} \]

### Step 7: Solve for potential energy \( U \)
Rearranging gives:
\[ U = -10\sqrt{3} \cdot \sqrt{3} = -10 \cdot 3 = -30 \, \text{J} \]

### Step 8: Final answer
Thus, the potential energy of the dipole is:
\[ U = -30 \, \text{J} \]