An electric dipole of length 20 cm having `+-3xx10^(-3)` C charge placed at `60^(@)` with respect to a uniform electric field experiences a torque of magnitude 6 Nm. The potential energy of the dipole is

Options

A. `-2sqrt3J`
B. `5sqrt3J`
C. `-3sqrt2J`
D. `3sqrt5J`

Explanation

To find the potential energy of the electric dipole, we will follow these steps:

### Step 1: Identify the given values
- Length of the dipole (2a) = 20 cm = 0.2 m
- Charge (Q) = ±3 x 10^(-3) C
- Angle (θ) = 60 degrees
- Torque (τ) = 6 Nm

### Step 2: Calculate the dipole moment (p)
The dipole moment (p) is given by the formula:
\[ p = Q \cdot 2a \]

Substituting the values:
\[ p = (3 \times 10^{-3} \, \text{C}) \cdot (0.2 \, \text{m}) = 6 \times 10^{-4} \, \text{C m} \]

### Step 3: Relate torque to electric field
The torque (τ) experienced by a dipole in an electric field (E) is given by:
\[ \tau = pE \sin \theta \]

Rearranging to find the electric field (E):
\[ E = \frac{\tau}{p \sin \theta} \]

### Step 4: Calculate the electric field (E)
Substituting the known values:
\[ E = \frac{6 \, \text{Nm}}{(6 \times 10^{-4} \, \text{C m}) \cdot \sin(60^\circ)} \]

Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\):
\[ E = \frac{6}{(6 \times 10^{-4}) \cdot \frac{\sqrt{3}}{2}} \]
\[ E = \frac{6 \cdot 2}{6 \times 10^{-4} \cdot \sqrt{3}} \]
\[ E = \frac{12}{6 \times 10^{-4} \cdot \sqrt{3}} = \frac{2}{10^{-4} \cdot \sqrt{3}} = \frac{2 \times 10^4}{\sqrt{3}} \]

### Step 5: Calculate the potential energy (U)
The potential energy (U) of a dipole in an electric field is given by:
\[ U = -pE \cos \theta \]

Substituting the values:
\[ U = - (6 \times 10^{-4} \, \text{C m}) \cdot \left(\frac{2 \times 10^4}{\sqrt{3}}\right) \cdot \cos(60^\circ) \]

Since \(\cos(60^\circ) = \frac{1}{2}\):
\[ U = - (6 \times 10^{-4}) \cdot \left(\frac{2 \times 10^4}{\sqrt{3}}\right) \cdot \frac{1}{2} \]
\[ U = - (6 \times 10^{-4}) \cdot \left(\frac{2 \times 10^4}{2\sqrt{3}}\right) \]
\[ U = - (6 \times 10^{-4}) \cdot \left(\frac{10^4}{\sqrt{3}}\right) \]
\[ U = - \frac{6 \times 10^0}{\sqrt{3}} \]
\[ U = -2\sqrt{3} \, \text{J} \]

### Final Answer
The potential energy of the dipole is:
\[ U = -2\sqrt{3} \, \text{J} \]

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