Question

Consider two identical dipoles \( D_1 \) and \( D_2 \). Charges \( -q \) and \( q \) of dipole \( D_1 \) are located at \( (0, 0) \) and \( (a, 0) \) and that of dipole \( D_2 \) at \( (0, a) \) and \( (0, 2a) \) in \( x \)-\( y \) plane, respectively. The net dipole moment of the system is

• A. \( qa(\hat{i}+\hat{j}) \)
• B. \( -qa(\hat{i}+\hat{j}) \)
• C. \( qa(\hat{i}-\hat{j}) \)
• D. \( -qa(\hat{i}-\hat{j}) \)

Hint:

Dipole moment points from the negative to the positive charge. D1 gives qa i and D2 gives qa j.

Answer 1

The Correct Option is A

Step 1: Recall the direction of a dipole moment:
The dipole moment of a pair of charges \( -q \) and \( +q \) is \( \vec{p} = q\,\vec{d} \). The vector \( \vec{d} \) points from the negative charge to the positive charge.
The net dipole moment of a system is the vector sum of the moments of its dipoles.

Step 2: Dipole D1:
\( -q \) is at \( (0,0) \) and \( +q \) is at \( (a,0) \). So \( \vec{d}_1 = a\hat{i} \).
Hence \( \vec{p}_1 = qa\,\hat{i} \).

Step 3: Dipole D2:
\( -q \) is at \( (0,a) \) and \( +q \) is at \( (0,2a) \). So \( \vec{d}_2 = (2a - a)\hat{j} = a\hat{j} \).
Hence \( \vec{p}_2 = qa\,\hat{j} \).

Step 4: Add the two moments:
\( \vec{p}_{net} = \vec{p}_1 + \vec{p}_2 = qa(\hat{i} + \hat{j}) \). This is option (A).
Option (B) has the wrong sign: it points from positive to negative charge. Options (C) and (D) have the wrong sign for the \( \hat{j} \) part, as if \( D_2 \) pointed down.

Final Answer:
\( \vec{p}_{net} = qa(\hat{i}+\hat{j}) \), option (A).