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 solve the problem of finding the electric field intensity ratio at points 4 m from the center of an electric dipole on the axial line and equatorial line, we can follow these steps:

### Given Data:
- Charge \( q_1 = +25 \times 10^{-9} \, \text{C} \)
- Charge \( q_2 = -25 \times 10^{-9} \, \text{C} \)
- Distance between the charges \( d = 6 \, \text{m} \)
- Distance from the center of the dipole to the point of interest \( r = 4 \, \text{m} \)

### Step 1: Calculate the Dipole Moment
The dipole moment \( p \) is given by:
\[ p = q \cdot d \]
where \( d \) is the distance between the charges. Here, \( d = 6 \, \text{m} \).

Since the dipole consists of two equal and opposite charges:
\[ p = (25 \times 10^{-9}) \times 6 = 150 \times 10^{-9} \, \text{C m} \]

### Step 2: Electric Field on the Axial Line
The electric field \( E_a \) at a distance \( r \) from the center of the dipole on the axial line is given by:
\[ E_a = \frac{2kp}{r^3} \]
where \( k \) is Coulomb's constant \( (8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) \).

Substituting the values:
\[ E_a = \frac{2 \times (8.99 \times 10^9) \times (150 \times 10^{-9})}{(4)^3} \]
Calculating \( (4)^3 = 64 \):
\[ E_a = \frac{2 \times (8.99 \times 10^9) \times (150 \times 10^{-9})}{64} \]
\[ E_a = \frac{2 \times 8.99 \times 150}{64} \times 10^0 \]
\[ E_a = \frac{2698.5}{64} \approx 42.13 \, \text{N/C} \]

### Step 3: Electric Field on the Equatorial Line
The electric field \( E_e \) at a distance \( r \) from the center of the dipole on the equatorial line is given by:
\[ E_e = \frac{kp}{(r^2 + a^2)^{3/2}} \]
where \( a \) is half the distance between the charges, \( a = \frac{d}{2} = 3 \, \text{m} \).

Substituting the values:
\[ E_e = \frac{(8.99 \times 10^9) \times (150 \times 10^{-9})}{(4^2 + 3^2)^{3/2}} \]
Calculating \( 4^2 + 3^2 = 16 + 9 = 25 \):
\[ E_e = \frac{(8.99 \times 10^9) \times (150 \times 10^{-9})}{(25)^{3/2}} \]
Calculating \( (25)^{3/2} = 125 \):
\[ E_e = \frac{(8.99 \times 10^9) \times (150 \times 10^{-9})}{125} \]
\[ E_e = \frac{1348.5}{125} \approx 10.788 \, \text{N/C} \]

### Step 4: Ratio of Electric Fields
Now, we find the ratio of the electric field intensities:
\[ \text{Ratio} = \frac{E_a}{E_e} = \frac{42.13}{10.788} \approx 3.90 \]

### Final Answer
Thus, the electric field intensity ratio at points 4 m from the center of the dipole on the axial line to the equatorial line is approximately:
\[ \text{Ratio} \approx 3.90 \]