The electric intensity due to a dipole of length `10 cm` and having a charge of `500 muC`, at a point on the axis at a distance `20 cm` from one of the charges in air is

Options

A. `6.25 xx 10^(7) N//C`
B. `9.28 xx 10^(7) N//C`
C. `13.1 xx 10^(11) N//C`
D. `20.5 xx 10^(7) N//C`

Answer

To find the electric intensity due to a dipole at a given point, we can follow these steps:

### Step 1: Identify the given values
- Length of the dipole, \( L = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Charge of each pole, \( q = 500 \, \mu\text{C} = 500 \times 10^{-6} \, \text{C} \)
- Distance from one of the charges to the point of interest, \( d = 20 \, \text{cm} = 0.2 \, \text{m} \)

### Step 2: Calculate the dipole moment \( P \)
The dipole moment \( P \) is calculated using the formula:
\[ P = 2q \cdot L \]
Substituting the values:
\[ P = 2 \times (500 \times 10^{-6}) \times 0.1 = 5 \times 10^{-5} \, \text{C m} \]

### Step 3: Calculate the distance \( R \) from the center of the dipole to the point
The distance \( R \) from the center of the dipole to the point on the axis is:
\[ R = d + \frac{L}{2} = 0.2 + 0.05 = 0.25 \, \text{m} \]

### Step 4: Use the formula for electric field \( E \) due to a dipole
The electric field \( E \) at a point on the axis of the dipole is given by:
\[ E = \frac{1}{4\pi \epsilon_0} \cdot \frac{2P}{R^3} \]
Where \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \) or \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

### Step 5: Substitute the values into the electric field formula
Using \( k \):
\[ E = k \cdot \frac{2P}{R^3} \]
Substituting \( P \) and \( R \):
\[ E = 9 \times 10^9 \cdot \frac{2 \cdot (5 \times 10^{-5})}{(0.25)^3} \]

### Step 6: Calculate \( R^3 \)
\[ R^3 = (0.25)^3 = 0.015625 \, \text{m}^3 \]

### Step 7: Calculate \( E \)
Substituting \( R^3 \) back into the equation:
\[ E = 9 \times 10^9 \cdot \frac{10 \times 10^{-5}}{0.015625} \]
Calculating the fraction:
\[ E = 9 \times 10^9 \cdot \frac{10^{-4}}{0.015625} = 9 \times 10^9 \cdot 6400 = 5.76 \times 10^7 \, \text{N/C} \]

### Step 8: Final result
Thus, the electric intensity due to the dipole at the specified point is approximately:
\[ E \approx 6.25 \times 10^7 \, \text{N/C} \]