A dipole is placed in xy plane parallel to the line `y = 2x`. There exists a uniform electric field along z-axis. Net force acting on the dipole will be zero. But it can experience some torque. We can show that the direction of this torque will be parallel to the line.

Options

A. `y = 2x + 1`
B. `y = -2x `
C. `y = -1/2x`
D. `y = -1/2 x + 2`

Explanation

To solve the problem step by step, we will analyze the given situation involving a dipole in an electric field.

### Step 1: Understand the Configuration
We have a dipole placed in the xy-plane, and it is aligned parallel to the line given by the equation \( y = 2x \). This means that the dipole moment vector \( \vec{p} \) can be expressed in terms of its components along the x and y axes.

### Step 2: Define the Dipole Moment
Since the dipole is along the line \( y = 2x \), we can express the dipole moment vector as:
\[ \vec{p} = p \hat{i} + 2p \hat{j} \]
where \( p \) is the magnitude of the dipole moment along the x-axis.

### Step 3: Define the Electric Field
The electric field \( \vec{E} \) is uniform and directed along the z-axis. We can represent it as:
\[ \vec{E} = E \hat{k} \]
where \( E \) is the magnitude of the electric field.

### Step 4: Calculate the Torque
The torque \( \vec{\tau} \) experienced by a dipole in an electric field is given by the cross product of the dipole moment and the electric field:
\[ \vec{\tau} = \vec{p} \times \vec{E} \]
Substituting the expressions for \( \vec{p} \) and \( \vec{E} \):
\[ \vec{\tau} = (p \hat{i} + 2p \hat{j}) \times (E \hat{k}) \]

### Step 5: Compute the Cross Product
Using the properties of the cross product:
\[ \vec{\tau} = pE (\hat{i} \times \hat{k}) + 2pE (\hat{j} \times \hat{k}) \]
From the right-hand rule, we know:
- \( \hat{i} \times \hat{k} = -\hat{j} \)
- \( \hat{j} \times \hat{k} = \hat{i} \)

Thus, we can write:
\[ \vec{\tau} = pE (-\hat{j}) + 2pE (\hat{i}) \]
This simplifies to:
\[ \vec{\tau} = 2pE \hat{i} - pE \hat{j} \]

### Step 6: Analyze the Direction of Torque
The torque vector can be expressed as:
\[ \vec{\tau} = (2pE) \hat{i} - (pE) \hat{j} \]
This indicates that the torque has components in both the x and y directions. The direction of the torque vector is parallel to the line defined by \( y = 2x \).

### Conclusion
The torque experienced by the dipole is not zero, and its direction is indeed parallel to the line \( y = 2x \).

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