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. `6.25xx10^(7)N//C`
B. `9.28 xx 10^(7)N//C`
C. `13.1xx11^(11)N//C`
D. `20.5xx10^(7)N//C`

Explanation

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

### Step 1: Understand the Problem
We have a dipole consisting of two charges, +q and -q, separated by a distance L. The dipole length is given as 10 cm, and the charge on each end is 500 µC. We need to find the electric field intensity at a point on the axis of the dipole, located 20 cm from one of the charges.

### Step 2: Convert Units
Convert all measurements to meters:
- Length of dipole, L = 10 cm = 0.1 m
- Distance from one charge to the point, r = 20 cm = 0.2 m

### Step 3: Calculate the Dipole Moment (P)
The dipole moment (P) is given by the formula:
\[ P = q \cdot L \]
Where:
- \( q = 500 \mu C = 500 \times 10^{-6} C = 5 \times 10^{-4} C \)
- \( L = 0.1 m \)

Calculating P:
\[ P = 5 \times 10^{-4} C \cdot 0.1 m = 5 \times 10^{-5} C \cdot m \]

### Step 4: Determine the Total Distance (R)
Since the distance is measured from one of the charges, we need to add half the dipole length to find the distance from the center of the dipole:
\[ R = 0.2 m + 0.05 m = 0.25 m \]

### Step 5: Use the Formula for Electric Field on the Axis of a Dipole
The electric field intensity (E) at a distance R from the center of the dipole on its axis is given by:
\[ E = \frac{1}{4 \pi \epsilon_0} \cdot \frac{2P}{R^3} \]
Where \( \epsilon_0 \) (the permittivity of free space) is approximately \( 8.85 \times 10^{-12} C^2/(N \cdot m^2) \).

Using \( k = \frac{1}{4 \pi \epsilon_0} \approx 9 \times 10^9 N \cdot m^2/C^2 \):
\[ E = k \cdot \frac{2P}{R^3} \]

### Step 6: Substitute Values into the Formula
Substituting the values we have:
- \( P = 5 \times 10^{-5} C \cdot m \)
- \( R = 0.25 m \)

\[ E = 9 \times 10^9 \cdot \frac{2 \cdot 5 \times 10^{-5}}{(0.25)^3} \]

### Step 7: Calculate R^3
Calculating \( R^3 \):
\[ R^3 = (0.25)^3 = 0.015625 m^3 \]

### Step 8: Calculate Electric Field Intensity
Now substituting back into the equation:
\[ E = 9 \times 10^9 \cdot \frac{10 \times 10^{-5}}{0.015625} \]
\[ E = 9 \times 10^9 \cdot \frac{10^{-4}}{0.015625} \]
\[ E = 9 \times 10^9 \cdot 6400 \]
\[ E = 5.76 \times 10^7 N/C \]

### Final Result
The electric field intensity due to the dipole at the given point is approximately:
\[ E \approx 6.25 \times 10^7 N/C \]