P and Q are two points on the axis and the perpendicular bisector respectively of an electric dipole. Both the points are far way from the dipole, and at equal distances from it. If `vecE_(P)` and `vecE_(Q)` are fields at P and Q ten

Options

A. `vecE_(P)=vecE_(Q)`
B. `vecE_(P)=-2vecE_(Q)`
C. `vecE_(P)=2vecE_(Q)`
D. `|E_(Q)=1/2|E_(P)|`, and `vecE_(Q)` is perpendicular to `vecE_(P)`

Answer

To solve the problem, we need to analyze the electric fields at points P and Q due to an electric dipole.

### Step-by-step Solution:

1. **Understanding the Configuration**:
- An electric dipole consists of two equal and opposite charges, +q and -q, separated by a distance 'd'.
- Point P is located on the axis of the dipole (the line extending through both charges), while point Q is located on the perpendicular bisector of the dipole.

2. **Distance from the Dipole**:
- Let the distance from the center of the dipole to points P and Q be 'r'. Given that both points are far away from the dipole, we can assume that \( r >> d \).

3. **Electric Field at Point P (On the Axis)**:
- The electric field \( \vec{E}_P \) at point P, which lies on the axis of the dipole, is given by the formula:
\[ \vec{E}_P = \frac{2kp}{r^3} \]
- Here, \( k \) is the electrostatic constant, and \( p \) is the dipole moment defined as \( p = q \cdot d \).
- The direction of \( \vec{E}_P \) is along the dipole moment (from +q to -q).

4. **Electric Field at Point Q (On the Perpendicular Bisector)**:
- The electric field \( \vec{E}_Q \) at point Q, which lies on the perpendicular bisector of the dipole, is given by the formula:
\[ \vec{E}_Q = \frac{kp}{r^3} \]
- The direction of \( \vec{E}_Q \) is opposite to the dipole moment.

5. **Relating \( \vec{E}_P \) and \( \vec{E}_Q \)**:
- From the expressions derived, we can relate \( \vec{E}_P \) and \( \vec{E}_Q \):
\[ \vec{E}_P = 2 \cdot \vec{E}_Q \]
- Since \( \vec{E}_Q \) is in the opposite direction of the dipole moment, we can express this as:
\[ \vec{E}_P = -2 \cdot \vec{E}_Q \]

### Final Result:
Thus, we conclude that:
\[ \vec{E}_P = -2 \cdot \vec{E}_Q \]