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

Explanation

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

### Step-by-Step Solution:

1. **Identify the Position of the Dipole and the Point**:
- The dipole is located at the origin (0, 0, 0) and has a dipole moment \( \vec{p} \) directed along the positive x-axis.
- The point of interest is (2, 2√2, 0).

2. **Calculate the Distance from the Dipole to the Point**:
- The distance \( R \) from the origin to the point (2, 2√2, 0) can be calculated using the distance formula:
\[ R = \sqrt{(2-0)^2 + (2\sqrt{2}-0)^2 + (0-0)^2} = \sqrt{4 + 8} = \sqrt{12} = 2\sqrt{3} \]

3. **Determine the Angle \( \theta \)**:
- The angle \( \theta \) between the dipole moment and the line connecting the dipole to the point can be found using the tangent function:
\[ \tan \theta = \frac{y}{x} = \frac{2\sqrt{2}}{2} = \sqrt{2} \]
- Therefore, \( \theta = \tan^{-1}(\sqrt{2}) \).

4. **Calculate the Components of the Electric Field**:
- The electric field \( \vec{E} \) due to a dipole at a point in space can be expressed in terms of the dipole moment and the angle:
\[ E = \frac{1}{4\pi \epsilon_0} \cdot \frac{2kp}{R^3} \]
- The components of the electric field can be calculated as:
- Along the dipole axis (x-direction):
\[ E_x = \frac{kp \cos \theta}{R^3} \]
- Perpendicular to the dipole axis (y-direction):
\[ E_y = \frac{kp \sin \theta}{R^3} \]

5. **Determine the Direction of the Electric Field**:
- Since the dipole moment is directed along the positive x-axis, the electric field will have a component in the negative x-direction (due to the nature of dipoles).
- The y-component will be in the positive y-direction.

6. **Result**:
- The net electric field will be directed in the y-direction since the y-component is positive and the x-component is negative. Thus, the direction of the electric field at the point (2, 2√2, 0) is along the positive y-axis.

### Final Answer:
The direction of the electric field at the point (2, 2√2, 0) is along the positive y-axis.

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