The electric field intensity at all points in space is given by `vecE = sqrt(3)hati - hatj` `V//m`. The nature of equipotential lines is x-y plane is given by

Options

A.
B.
C.
D.

Answer

To solve the problem, we need to analyze the given electric field vector and determine the nature of the equipotential lines in the x-y plane.

### Step-by-Step Solution:

1. **Identify the Electric Field Vector**:
The electric field intensity is given as:
\[ \vec{E} = \sqrt{3} \hat{i} - \hat{j} \quad \text{(V/m)} \]
Here, we can identify the components of the electric field:
- \( E_x = \sqrt{3} \)
- \( E_y = -1 \)

2. **Calculate the Angle of the Electric Field**:
The angle \( \theta \) that the electric field makes with the x-axis can be calculated using the tangent function:
\[ \theta = \tan^{-1}\left(\frac{E_y}{E_x}\right) = \tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) \]
Evaluating this gives:
\[ \theta = -30^\circ \]

3. **Determine the Angle of the Equipotential Lines**:
Equipotential lines are always perpendicular to the electric field lines. Therefore, the angle \( \phi \) that the equipotential lines make with the x-axis is given by:
\[ \phi = 90^\circ + \theta \]
Substituting the value of \( \theta \):
\[ \phi = 90^\circ - 30^\circ = 60^\circ \]

4. **Direction of Potential Change**:
The electric field points in the direction of decreasing potential. Since \( E_x \) is positive and \( E_y \) is negative, the potential decreases in the positive x-direction and negative y-direction. This indicates that the equipotential lines will be oriented such that they decrease in potential towards the fourth quadrant.

5. **Conclusion**:
The equipotential lines will be straight lines making an angle of \( 60^\circ \) with the x-axis, and they will be oriented such that the potential decreases towards the fourth quadrant.

### Final Answer:
The nature of the equipotential lines in the x-y plane is that they are straight lines making an angle of \( 60^\circ \) with the x-axis, decreasing in potential towards the fourth quadrant.

---