Electric charges `q,q,2q` are placed at the equilateral triangle ABC of side l .The electric dipole moment of the system is ?

Options

A. `sqrt(3)ql`
B. `sqrt(2)ql`
C. ql
D. `2ql`

Explanation

To find the electric dipole moment of the system with charges \( q, q, \) and \( 2q \) placed at the vertices of an equilateral triangle ABC with side length \( l \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions**:
- Let the charges be placed as follows:
- Charge \( q \) at vertex A,
- Charge \( q \) at vertex B,
- Charge \( 2q \) at vertex C.
- The vertices A, B, and C form an equilateral triangle with side length \( l \).

2. **Calculate the Dipole Moment for Each Pair**:
- The electric dipole moment \( \vec{p} \) is defined as:
\[ \vec{p} = q \cdot \vec{d} \]
where \( \vec{d} \) is the vector pointing from the negative charge to the positive charge.
- We will calculate the dipole moments for the pairs (A, C) and (B, C).

3. **Calculate Dipole Moment \( \vec{p_1} \) for Pair (A, C)**:
- For the pair (A, C):
- Charge at A is \( q \) (positive),
- Charge at C is \( 2q \) (negative).
- The distance \( d \) between A and C is \( l \).
- The dipole moment \( \vec{p_1} \) is given by:
\[ \vec{p_1} = 2q \cdot \frac{l}{2} = ql \quad \text{(acting from C to A)} \]

4. **Calculate Dipole Moment \( \vec{p_2} \) for Pair (B, C)**:
- For the pair (B, C):
- Charge at B is \( q \) (positive),
- Charge at C is \( 2q \) (negative).
- The distance \( d \) between B and C is also \( l \).
- The dipole moment \( \vec{p_2} \) is given by:
\[ \vec{p_2} = 2q \cdot \frac{l}{2} = ql \quad \text{(acting from C to B)} \]

5. **Resultant Dipole Moment**:
- The angle between \( \vec{p_1} \) and \( \vec{p_2} \) is \( 60^\circ \) because of the equilateral triangle.
- Using the formula for the resultant of two vectors:
\[ p_{\text{net}} = \sqrt{p_1^2 + p_2^2 + 2p_1p_2 \cos(60^\circ)} \]
- Since \( p_1 = p_2 = p \):
\[ p_{\text{net}} = \sqrt{p^2 + p^2 + 2p^2 \cdot \frac{1}{2}} = \sqrt{2p^2 + p^2} = \sqrt{3p^2} = \sqrt{3}p \]

6. **Substituting the Value of \( p \)**:
- We have \( p = ql \), so:
\[ p_{\text{net}} = \sqrt{3} \cdot ql \]

### Final Answer:
The electric dipole moment of the system is:
\[ \vec{p}_{\text{net}} = \sqrt{3}ql \]