Electric field on the axis of a small electric dipole at a distance r is `vec(E)_(1)` and `vec(E)_(2)` at a distance of `2r` on a line of perpendicular bisector is

Options

A. `E_(2)=-E_(1)//8`
B. `E_(2)=-E_(1)//16`
C. `E_(2)=-E_(1)//4`
D. `E_(2)=E_(1)//8`

Answer

To solve the problem of finding the electric field on the axis of a small electric dipole at a distance \( r \) (denoted as \( \vec{E}_1 \)) and at a distance \( 2r \) on the line of the perpendicular bisector (denoted as \( \vec{E}_2 \)), we will follow these steps:

### Step 1: Understand the Electric Dipole
An electric dipole consists of two equal and opposite charges, \( +q \) and \( -q \), separated by a distance \( d \). The electric dipole moment \( \vec{p} \) is defined as:
\[ \vec{p} = q \cdot d \]

### Step 2: Calculate the Electric Field on the Axis of the Dipole
The electric field \( \vec{E}_1 \) at a distance \( r \) along the axis of the dipole is given by the formula:
\[ \vec{E}_1 = \frac{2kp}{r^3} \]
where \( k \) is Coulomb's constant and \( p \) is the dipole moment.

### Step 3: Calculate the Electric Field on the Perpendicular Bisector
The electric field \( \vec{E}_2 \) at a distance \( 2r \) on the perpendicular bisector of the dipole is given by:
\[ \vec{E}_2 = \frac{kp}{(2r)^3} = \frac{kp}{8r^3} \]

### Step 4: Compare the Magnitudes of \( \vec{E}_1 \) and \( \vec{E}_2 \)
Now, we can find the ratio of the magnitudes of \( \vec{E}_1 \) and \( \vec{E}_2 \):
\[ \frac{|\vec{E}_1|}{|\vec{E}_2|} = \frac{\frac{2kp}{r^3}}{\frac{kp}{8r^3}} = \frac{2kp \cdot 8r^3}{kp \cdot r^3} = 16 \]
Thus, we have:
\[ |\vec{E}_1| = 16 |\vec{E}_2| \]

### Step 5: Determine the Direction of the Electric Fields
Since \( \vec{E}_1 \) is directed along the dipole moment and \( \vec{E}_2 \) is directed opposite to the dipole moment, we can express this relationship as:
\[ \vec{E}_2 = -\frac{1}{16} \vec{E}_1 \]

### Final Answer
Thus, the electric field on the axis of the dipole at a distance \( r \) is \( \vec{E}_1 \) and at a distance \( 2r \) on the perpendicular bisector is \( \vec{E}_2 \), where:
\[ \vec{E}_2 = -\frac{1}{16} \vec{E}_1 \]