Two short dipoles, each of diple moment p , are placed at origin. The dipole moment of one dipole is along x-axis, while that of other is along y-axis . The electric field at a point (a,0) is given by

Options

A. `((1)/( 4pi epsilon_(0))) (2p)/(a^(3))`
B. `((1)/( 4pi epsilon_(0))) (p)/(a^(3))`
C. `((1)/( 4pi epsilon_(0))) (sqrt(5)p)/(a^(3))`
D. Zero

Answer

To find the electric field at the point (a, 0) due to two dipoles placed at the origin, we will follow these steps:

### Step 1: Understanding the Dipoles
We have two dipoles, each with a dipole moment \( \vec{p} \). One dipole is oriented along the x-axis, and the other is oriented along the y-axis.

### Step 2: Electric Field Due to the Dipole Along the X-axis
The electric field \( \vec{E}_x \) at a point due to a dipole along the x-axis at a distance \( r \) is given by:
\[ \vec{E}_x = \frac{2kp}{r^3} \hat{i} \]
where \( k \) is the Coulomb's constant, \( p \) is the dipole moment, and \( \hat{i} \) is the unit vector in the x-direction.

At the point (a, 0), the distance \( r \) is \( a \), so:
\[ \vec{E}_x = \frac{2kp}{a^3} \hat{i} \]

### Step 3: Electric Field Due to the Dipole Along the Y-axis
The electric field \( \vec{E}_y \) at the same point (a, 0) due to the dipole along the y-axis is given by:
\[ \vec{E}_y = -\frac{kp}{r^3} \hat{j} \]
Here, the electric field is directed opposite to the dipole moment because the point (a, 0) is at the equatorial position of the dipole along the y-axis.

Again, substituting \( r = a \):
\[ \vec{E}_y = -\frac{kp}{a^3} \hat{j} \]

### Step 4: Net Electric Field
The net electric field \( \vec{E}_{net} \) at the point (a, 0) is the vector sum of \( \vec{E}_x \) and \( \vec{E}_y \):
\[ \vec{E}_{net} = \vec{E}_x + \vec{E}_y = \frac{2kp}{a^3} \hat{i} - \frac{kp}{a^3} \hat{j} \]

### Step 5: Simplifying the Expression
We can factor out \( \frac{kp}{a^3} \):
\[ \vec{E}_{net} = \frac{kp}{a^3} \left( 2 \hat{i} - \hat{j} \right) \]

### Step 6: Magnitude of the Electric Field
To find the magnitude of the electric field:
\[ |\vec{E}_{net}| = \frac{kp}{a^3} \sqrt{(2)^2 + (-1)^2} = \frac{kp}{a^3} \sqrt{4 + 1} = \frac{kp}{a^3} \sqrt{5} \]

### Step 7: Final Expression
Substituting \( k \) with \( \frac{1}{4 \pi \epsilon_0} \):
\[ |\vec{E}_{net}| = \frac{1}{4 \pi \epsilon_0} \cdot \frac{\sqrt{5} p}{a^3} \]

Thus, the electric field at the point (a, 0) is given by:
\[ \vec{E}_{net} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{\sqrt{5} p}{a^3} \hat{r} \]
where \( \hat{r} \) is the direction of the resultant electric field.