Three point charges +q, –2q and +q are placed at points (x = 0, y = a, z = 0), (x = 0,y = 0, z = 0) and (x = a, y = 0, z = 0) respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are:-

Options

A. `sqrt(2)` qa along + x direction
B. `sqrt(2)` qa along + y direction
C. `sqrt(2)` qa along the line joining points (x=0,y=0,z=0) and (x=a,y=a,z=0)
D. qa along the line joining points (x=0,y=0,z=0) and (z=a , y=a ,z=0)

Explanation

To find the electric dipole moment vector of the given charge assembly, we can follow these steps:

### Step 1: Identify the positions and charges
We have three point charges:
- Charge \( +q \) at position \( (0, a, 0) \)
- Charge \( -2q \) at position \( (0, 0, 0) \)
- Charge \( +q \) at position \( (a, 0, 0) \)

### Step 2: Calculate the dipole moment contributions
The electric dipole moment \( \vec{p} \) for a system of point charges is given by the formula:
\[ \vec{p} = \sum q_i \vec{r}_i \]
where \( q_i \) is the charge and \( \vec{r}_i \) is the position vector of the charge.

#### Contribution from \( +q \) at \( (0, a, 0) \):
\[ \vec{p_1} = +q \cdot (0, a, 0) = (0, qa, 0) \]

#### Contribution from \( -2q \) at \( (0, 0, 0) \):
\[ \vec{p_2} = -2q \cdot (0, 0, 0) = (0, 0, 0) \]

#### Contribution from \( +q \) at \( (a, 0, 0) \):
\[ \vec{p_3} = +q \cdot (a, 0, 0) = (qa, 0, 0) \]

### Step 3: Sum the contributions
Now, we sum the contributions from all three charges:
\[ \vec{p} = \vec{p_1} + \vec{p_2} + \vec{p_3} = (0, qa, 0) + (0, 0, 0) + (qa, 0, 0) = (qa, qa, 0) \]

### Step 4: Write the dipole moment vector
Thus, the total electric dipole moment vector is:
\[ \vec{p} = (qa, qa, 0) = qa \hat{i} + qa \hat{j} \]

### Step 5: Calculate the magnitude of the dipole moment
The magnitude of the dipole moment \( |\vec{p}| \) is given by:
\[ |\vec{p}| = \sqrt{(qa)^2 + (qa)^2} = \sqrt{2(qa)^2} = qa\sqrt{2} \]

### Step 6: Determine the direction of the dipole moment
The direction of the dipole moment vector \( \vec{p} \) can be expressed in terms of angles. Since both components are equal, the angle with respect to the axes can be determined using:
\[ \tan \theta = \frac{p_y}{p_x} = \frac{qa}{qa} = 1 \implies \theta = 45^\circ \]

### Final Result
The magnitude of the electric dipole moment vector is \( qa\sqrt{2} \) and the direction is at an angle of \( 45^\circ \) from both the x-axis and y-axis.