The magnitude of electric field intensity at point `B(2,0,0)` due to dipole of dipole moment, `vec(p)=hat(i)+sqrt(3)hat(j)` kept at origin is (assume that the point `B` is at large distance from the dipole and `k=(1)/(4pi epsilon_(0)))`

Options

A. `(sqrt(13)K)/8`
B. `(sqrt(13)K)/4`
C. `(sqrt(7)K)/8`
D. `(sqrt(7)K)/4`

Answer

To find the magnitude of the electric field intensity at point \( B(2,0,0) \) due to a dipole moment \( \vec{p} = \hat{i} + \sqrt{3} \hat{j} \) located at the origin, we can follow these steps:

### Step 1: Identify the dipole moment and its components
The dipole moment is given as:
\[ \vec{p} = \hat{i} + \sqrt{3} \hat{j} \]
This can be expressed in terms of its components:
- \( p_x = 1 \) (coefficient of \( \hat{i} \))
- \( p_y = \sqrt{3} \) (coefficient of \( \hat{j} \))

### Step 2: Calculate the magnitude of the dipole moment \( p \)
The magnitude of the dipole moment \( p \) is given by:
\[ p = \sqrt{p_x^2 + p_y^2} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

### Step 3: Determine the position vector from the dipole to point \( B \)
The position vector \( \vec{r} \) from the dipole (at the origin) to point \( B(2,0,0) \) is:
\[ \vec{r} = 2 \hat{i} + 0 \hat{j} + 0 \hat{k} \]
The distance \( R \) from the dipole to point \( B \) is:
\[ R = |\vec{r}| = 2 \]

### Step 4: Calculate the angle \( \theta \) between the dipole moment and the position vector
To find the angle \( \theta \) between the dipole moment \( \vec{p} \) and the position vector \( \vec{r} \), we can use the dot product:
\[ \cos \theta = \frac{\vec{p} \cdot \vec{r}}{|\vec{p}| |\vec{r}|} \]
Calculating the dot product:
\[ \vec{p} \cdot \vec{r} = (1)(2) + (\sqrt{3})(0) = 2 \]
The magnitudes are:
- \( |\vec{p}| = 2 \)
- \( |\vec{r}| = 2 \)

Thus,
\[ \cos \theta = \frac{2}{2 \cdot 2} = \frac{1}{2} \]
This gives us:
\[ \theta = 60^\circ \]

### Step 5: Use the formula for the electric field due to a dipole
The formula for the electric field \( E \) at a point making an angle \( \theta \) with the dipole moment is:
\[ E = \frac{k p}{R^3} \left( 1 + 3 \cos^2 \theta \right) \]
Substituting the known values:
- \( k = \frac{1}{4 \pi \epsilon_0} \)
- \( p = 2 \)
- \( R = 2 \)
- \( \cos 60^\circ = \frac{1}{2} \)

Calculating \( E \):
\[ E = \frac{k \cdot 2}{2^3} \left( 1 + 3 \left( \frac{1}{2} \right)^2 \right) \]
\[ E = \frac{k \cdot 2}{8} \left( 1 + 3 \cdot \frac{1}{4} \right) \]
\[ E = \frac{k \cdot 2}{8} \left( 1 + \frac{3}{4} \right) = \frac{k \cdot 2}{8} \cdot \frac{7}{4} \]
\[ E = \frac{k \cdot 2 \cdot 7}{32} = \frac{7k}{16} \]

### Step 6: Final expression for the electric field
Thus, the magnitude of the electric field intensity at point \( B(2,0,0) \) is:
\[ E = \frac{7k}{16} \]