An electric dipole has a fixed dipole moment `vec(p)`, which makes angle `theta` with respect to x-axis. When subjected to an electric field `vec(E_(1)) = E hat(i)`, it experiences a torque `vec(T_(1)) = tau hat(k)`. When subjected to another electric field `vec(E_(2)) = sqrt(3) E_(1) hat(j)`, it experiences torque `vec(T_(2)) = -vec(T_(1))`. The angle `theta` is

Options

A. `434J/kg-""^(@)C`
B. `378J/kg-""^(@)C`
C. `315J/kg-""^(@)C`
D. `476J/kg-""^(@)C`

Answer

To solve the problem, we need to analyze the torques experienced by the electric dipole in two different electric fields. Let's break it down step by step.

### Step 1: Define the Dipole Moment and Electric Fields
The dipole moment \(\vec{p}\) can be expressed in terms of its components:
\[ \vec{p} = p \cos \theta \hat{i} + p \sin \theta \hat{j} \]
where \(p\) is the magnitude of the dipole moment and \(\theta\) is the angle it makes with the x-axis.

The electric fields are given as:
\[ \vec{E_1} = E \hat{i} \]
\[ \vec{E_2} = \sqrt{3} E \hat{j} \]

### Step 2: Calculate the Torque in the First Electric Field
The torque \(\vec{T_1}\) experienced by the dipole in the electric field \(\vec{E_1}\) is given by the formula:
\[ \vec{T_1} = \vec{p} \times \vec{E_1} \]
Substituting the values:
\[ \vec{T_1} = (p \cos \theta \hat{i} + p \sin \theta \hat{j}) \times (E \hat{i}) \]
Using the properties of the cross product:
\[ \hat{i} \times \hat{i} = 0 \quad \text{and} \quad \hat{j} \times \hat{i} = -\hat{k} \]
Thus, we have:
\[ \vec{T_1} = p \sin \theta E (-\hat{k}) = -p E \sin \theta \hat{k} \]

### Step 3: Calculate the Torque in the Second Electric Field
Now, we calculate the torque \(\vec{T_2}\) in the electric field \(\vec{E_2}\):
\[ \vec{T_2} = \vec{p} \times \vec{E_2} \]
Substituting the values:
\[ \vec{T_2} = (p \cos \theta \hat{i} + p \sin \theta \hat{j}) \times (\sqrt{3} E \hat{j}) \]
Using the properties of the cross product:
\[ \hat{i} \times \hat{j} = \hat{k} \quad \text{and} \quad \hat{j} \times \hat{j} = 0 \]
Thus, we have:
\[ \vec{T_2} = p \cos \theta \sqrt{3} E \hat{k} \]

### Step 4: Relate the Two Torques
According to the problem, we know that:
\[ \vec{T_2} = -\vec{T_1} \]
This gives us:
\[ p \cos \theta \sqrt{3} E \hat{k} = p E \sin \theta \hat{k} \]
Removing the common factors \(pE\) (assuming \(p\) and \(E\) are not zero), we get:
\[ \cos \theta \sqrt{3} = -\sin \theta \]

### Step 5: Solve for \(\theta\)
Rearranging the equation:
\[ \sqrt{3} \cos \theta + \sin \theta = 0 \]
Dividing through by \(\cos \theta\):
\[ \sqrt{3} + \tan \theta = 0 \]
Thus:
\[ \tan \theta = -\sqrt{3} \]
This implies:
\[ \theta = \tan^{-1}(-\sqrt{3}) \]
The angle \(\theta\) that satisfies this is:
\[ \theta = -\frac{\pi}{3} \text{ or } \theta = \frac{2\pi}{3} \text{ (in the second quadrant)} \]
However, since we are looking for the angle with respect to the x-axis, we can conclude:
\[ \theta = \frac{2\pi}{3} \text{ or } 120^\circ \]

### Final Answer
\[ \theta = 120^\circ \]