A small electric dipole is placed at origin with its dipole moment directed along positive x-axis .The direction of electric field at point `(2,2sqrt(2),0)`

Options

A. along z-aixs
B. along y-axis
C. along negative y-axis
D. along negative z-axis

Answer

To determine the direction of the electric field at the point (2, 2√2, 0) due to an electric dipole placed at the origin with its dipole moment directed along the positive x-axis, we can follow these steps:

### Step 1: Identify the Position Vector
The position vector \( \mathbf{r} \) of the point where we want to find the electric field is given as \( (2, 2\sqrt{2}, 0) \).

### Step 2: Calculate the Angle \( \theta \)
The angle \( \theta \) between the dipole moment (along the x-axis) and the position vector can be calculated using the tangent function:
\[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{2\sqrt{2}}{2} = \sqrt{2} \]
From this, we can find \( \theta \):
\[ \theta = \tan^{-1}(\sqrt{2}) \]

### Step 3: Components of the Electric Field
The electric field \( \mathbf{E} \) due to a dipole at a point in space can be resolved into two components: radial and tangential. The radial component is directed along the line connecting the dipole and the point, while the tangential component is perpendicular to it.

### Step 4: Determine the Resultant Electric Field
Let’s denote the angle made by the resultant electric field with the x-axis as \( \alpha \). The relationship between the angles can be established using the tangent addition formula:
\[ \tan(\alpha + \theta) = \frac{\tan \alpha + \tan \theta}{1 - \tan \alpha \tan \theta} \]
Given that \( \tan \theta = \sqrt{2} \) and assuming \( \tan \alpha = \frac{1}{\sqrt{2}} \) (as derived from the geometry of the problem), we can substitute these values into the formula.

### Step 5: Simplify the Expression
Substituting \( \tan \alpha = \frac{1}{\sqrt{2}} \) and \( \tan \theta = \sqrt{2} \):
\[ \tan(\alpha + \theta) = \frac{\frac{1}{\sqrt{2}} + \sqrt{2}}{1 - \left(\frac{1}{\sqrt{2}} \cdot \sqrt{2}\right)} = \frac{\frac{1 + 2}{\sqrt{2}}}{1 - 1} = \text{undefined} \]
This indicates that \( \alpha + \theta = 90^\circ \).

### Step 6: Conclusion
Since \( \alpha + \theta = 90^\circ \), the resultant electric field is perpendicular to the x-axis, which means it is directed along the y-axis.

Thus, the direction of the electric field at the point \( (2, 2\sqrt{2}, 0) \) is along the positive y-axis.

### Final Answer
The direction of the electric field at the point \( (2, 2\sqrt{2}, 0) \) is along the positive y-axis.

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