Electric field_ on the axis of a small electric dipole at a distance r is `bar(E_1)` and_ at a distance of 2r on a line of perpendicular bisector is `bar(E_2)`. Then

Options

A. `bar(E_2)= - (bar(E_1))/(8)`
B. `bar(E_2) = (bar(E_1))/(16)`
C. `bar(E_2)= -(bar(E_1))/(4)`
D. `bar(E_2)=-(bar(E_1))/(10)`

Answer

To solve the problem, we need to find the relationship between the electric fields \( \bar{E_1} \) and \( \bar{E_2} \) at specified distances from a small electric dipole.

### Step-by-Step Solution:

1. **Understanding the Electric Dipole**:
An electric dipole consists of two equal and opposite charges separated by a small distance. The dipole moment \( p \) is defined as \( p = q \cdot d \), where \( q \) is the charge and \( d \) is the separation between the charges.

2. **Electric Field on the Axis of the Dipole**:
The electric field \( \bar{E_1} \) at a distance \( r \) along the axis of a dipole is given by the formula:
\[ \bar{E_1} = \frac{2kp}{r^3} \hat{i} \]
where \( k \) is Coulomb's constant and \( \hat{i} \) is the unit vector in the direction of the dipole moment.

3. **Electric Field on the Perpendicular Bisector**:
The electric field \( \bar{E_2} \) at a distance \( 2r \) on the perpendicular bisector of the dipole is given by:
\[ \bar{E_2} = -\frac{kp}{(2r)^3} \hat{i} = -\frac{kp}{8r^3} \hat{i} \]
The negative sign indicates that the direction of the electric field is opposite to the dipole moment direction.

4. **Finding the Ratio \( \frac{\bar{E_2}}{\bar{E_1}} \)**:
Now we can find the ratio of \( \bar{E_2} \) to \( \bar{E_1} \):
\[ \frac{\bar{E_2}}{\bar{E_1}} = \frac{-\frac{kp}{8r^3} \hat{i}}{\frac{2kp}{r^3} \hat{i}} \]
Simplifying this expression:
\[ = \frac{-\frac{1}{8}}{\frac{2}{1}} = -\frac{1}{16} \]

5. **Final Result**:
Therefore, we can write:
\[ \bar{E_2} = -\frac{1}{16} \bar{E_1} \]

### Conclusion:
The relationship between the electric fields is:
\[ \bar{E_2} = -\frac{1}{16} \bar{E_1} \]