Two charges of `+25xx10^(-9)` coulomb and `-25xx10^(-9)` coulomb are placed 6 m apart. Find the electric field intensity ratio at points 4 m from the centre of the electric dipole (i) on axial line (ii) on equatorial line

Options

A. `(1000)/(49)`
B. `(49)/(1000)`
C. `(500)/(49)`
D. `(49)/(500)`

Answer

To find the electric field intensity ratio at points 4 m from the center of the electric dipole (i) on the axial line and (ii) on the equatorial line, we can follow these steps:

### Step 1: Understand the Configuration
We have two charges:
- \( q_1 = +25 \times 10^{-9} \) C (positive charge)
- \( q_2 = -25 \times 10^{-9} \) C (negative charge)

The distance between the two charges is 6 m. The center of the dipole is at the midpoint between the two charges, which is 3 m from each charge.

### Step 2: Define the Points of Interest
- **Axial Line:** A point \( P \) is located 4 m from the center of the dipole along the axial line.
- **Equatorial Line:** A point \( Q \) is also located 4 m from the center but on the equatorial line.

### Step 3: Calculate Electric Field Intensity on the Axial Line
The formula for the electric field intensity \( E_a \) on the axial line of a dipole is given by:
\[ E_a = \frac{2kp}{r^2 - a^2} \]
where:
- \( k \) is Coulomb's constant (\( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \))
- \( p \) is the dipole moment, given by \( p = q \cdot d \) (where \( d \) is the distance between the charges)
- \( r \) is the distance from the center to the point of interest (4 m)
- \( a \) is half the distance between the charges (3 m)

Calculating the dipole moment \( p \):
\[ p = q \cdot d = (25 \times 10^{-9}) \cdot 6 = 150 \times 10^{-9} \, \text{C m} \]

Now substituting values into the axial electric field formula:
\[ E_a = \frac{2 \cdot (9 \times 10^9) \cdot (150 \times 10^{-9})}{4^2 - 3^2} \]
Calculating the denominator:
\[ 4^2 - 3^2 = 16 - 9 = 7 \]
Thus,
\[ E_a = \frac{2 \cdot (9 \times 10^9) \cdot (150 \times 10^{-9})}{7} \]

### Step 4: Calculate Electric Field Intensity on the Equatorial Line
The formula for the electric field intensity \( E_e \) on the equatorial line of a dipole is given by:
\[ E_e = \frac{kp}{r^2 + a^2} \]
Substituting the same values:
\[ E_e = \frac{(9 \times 10^9) \cdot (150 \times 10^{-9})}{4^2 + 3^2} \]
Calculating the denominator:
\[ 4^2 + 3^2 = 16 + 9 = 25 \]
Thus,
\[ E_e = \frac{(9 \times 10^9) \cdot (150 \times 10^{-9})}{25} \]

### Step 5: Calculate the Ratio of Electric Fields
Now, we find the ratio \( \frac{E_a}{E_e} \):
\[ \frac{E_a}{E_e} = \frac{\frac{2 \cdot (9 \times 10^9) \cdot (150 \times 10^{-9})}{7}}{\frac{(9 \times 10^9) \cdot (150 \times 10^{-9})}{25}} \]
The \( (9 \times 10^9) \) and \( (150 \times 10^{-9}) \) cancel out:
\[ \frac{E_a}{E_e} = \frac{2 \cdot 25}{7} = \frac{50}{7} \]

### Final Result
Thus, the electric field intensity ratio at the two points is:
\[ \frac{E_a}{E_e} = \frac{1000}{49} \]