A short electric dipoleis placed along y-axis at the origin. The electric field vector at a point P on x-axis is `vec(E ) _(1)`. The dipole is rotated by `( pi)/(2)` at its positin . Now, the electric field vector at point P is `vec(E )_(2)`. Select the correct alternative(s) `:`

Options

A. `vec(E )_(2)= - vec(E )_(1)`
B. `vec(E )_(2)= - 2vec(E )_(1)`
C. `|vec(E )_(2)| = - 2|vec(E )_(1)|`
D. `vec(E )_(2) . vec( E )_(1)=0`

Answer

To solve the problem, we need to analyze the electric field produced by a short electric dipole at a point on the x-axis before and after the dipole is rotated by \( \frac{\pi}{2} \).

### Step-by-Step Solution:

**Step 1: Understand the Configuration of the Dipole**
- The electric dipole consists of two charges, +q and -q, separated by a small distance \( d \).
- Initially, the dipole is oriented along the y-axis, which means the dipole moment \( \vec{p} \) points in the positive y-direction.

**Step 2: Determine the Electric Field at Point P (E1)**
- Point P is located on the x-axis at a distance \( d \) from the origin.
- Since the dipole is along the y-axis, point P is an equatorial point with respect to the dipole.
- The electric field \( \vec{E}_1 \) at point P due to the dipole can be calculated using the formula for the electric field at an equatorial point:
\[ \vec{E}_1 = -\frac{k \cdot p}{d^3} \]
where \( k \) is a constant related to the electric field.

**Step 3: Rotate the Dipole by \( \frac{\pi}{2} \)**
- When the dipole is rotated by \( \frac{\pi}{2} \), it now points along the x-axis.
- The dipole moment \( \vec{p} \) now points in the positive x-direction.

**Step 4: Determine the Electric Field at Point P (E2)**
- After the rotation, point P is now an axial point with respect to the dipole.
- The electric field \( \vec{E}_2 \) at point P can be calculated using the formula for the electric field at an axial point:
\[ \vec{E}_2 = \frac{2k \cdot p}{d^3} \]

**Step 5: Analyze the Relationship Between E1 and E2**
- From the calculations:
- \( \vec{E}_1 = -\frac{k \cdot p}{d^3} \) (pointing in the negative x-direction)
- \( \vec{E}_2 = \frac{2k \cdot p}{d^3} \) (pointing in the positive x-direction)
- Since \( \vec{E}_1 \) and \( \vec{E}_2 \) are in opposite directions, they are perpendicular to each other.

**Step 6: Conclusion**
- The dot product of two perpendicular vectors is zero:
\[ \vec{E}_1 \cdot \vec{E}_2 = |\vec{E}_1| |\vec{E}_2| \cos(90^\circ) = 0 \]
- Therefore, the correct alternative is that \( \vec{E}_1 \cdot \vec{E}_2 = 0 \).